#### Demystifying Permutation & Combination - Rajesh Balasubramanian - CAT 100th Percentile - CAT 2011, 2012 and 2014 (Part 1/7)

Permutation and Combination is a wonderful topic to learn. This topic is all about counting. In all other topics, we need to get to an answer, be it average, or speed, or angle, or LCM. In this topic, we will mostly concern ourselves with the idea of counting the number of ways of doing something. In most scenarios we will be enumerating possibilities rather than solving to get an answer. Before we go further, let us define a few ground rules.

- This topic is based on the simple idea of counting. So, from now on, we will avoid the terms ‘permutation’ and ‘combination’.
- We will build the theory for this topic mostly with examples.
- Wherever possible, we will list and count. Especially, if the answer is less than 10 we will shoot to list out all possibilities.
- nCr, nPr, n! will be resorted to only if absolutely necessary. Counting errors happen when we look to force–fit an idea on to a question. We will use nPr, nCr etc, when we hit the framework. We will not use these as starting points.

Now let us move to first set of questions.

**Ram wants to travel from Chennai to Kolkatta (to join IIM Kolkatta). He wants to take only 1 piece of baggage with him. He has 3 types of suitcases and 4 types of back packs. In how many ways can he select his luggage?**

This is straightforward. Out of the seven pieces available, he has to select exactly one. If he has suitcases S1, S2, S3 and bags B1, B2, B3 and B4, he can pick one of these 7. So, there are 7 options.

After trying to fit in his luggage, Ram realizes that he needs to carry two pieces of baggage. He plans to carry one suitcase and one backpack. In how many ways can he select his baggage now?

He can select S1 or S2 or S3 and B1, B2, B3 or B4. Totally he has 12 options now.

S1B1, S1B2, S1B3, S1B4

S2B1, S2B2, S2B3, S2B4

S3B1, S3B2, S3B3, S3B4

**Fundamental Rule of Counting**

If there are ‘m’ ways of doing a task and ‘n’ ways of doing another task, then there are m × n ways of doing both.

If a process can be broken into multiple steps and you have to do each of the many steps, then the total number of ways of doing the process = product of number of ways of doing each step.

AND => ×

**Rule of Sum**

If there are m ways of doing a task and n ways of doing another task, then the number of ways of doing either of these two (but not both) = m + n. In other words, if we have to do one thing OR another thing, the number of ways = sum of the number of ways of doing each step

OR => +

Number of ways Ram can choose 1 suitcase = 3

Number of ways Ram can choose 1 bag = 4

Number of ways he can choose 1 suitcase OR 1

bag = 3 + 4 = 7

Number of ways he can choose 1 suitcase AND 1 bag = 3 × 4 = 12

Let us take one more example and build counting ideas.

**John, who is in charge of creating number plates for cars, has three colors at his disposal – black, white and yellow. He has to paint the number plate with one color and the letters and numbers in another color. In how many ways can he create a number plate?**

Again, let us list and count.

- Black plate, white letters
- Black plate, yellow letters
- White plate, yellow letters
- White plate, black letters
- Yellow plate, white letters
- Yellow plate, black letters

There are totally six ways.

**Mike has three types of flowers with him – roses, lilies and violets. Mike decides he has to create a bouquet that has two distinct flowers. In how many ways can he create this?**

- Rose and Lily
- Lily and Violet
- Rose and Violet

What is the difference?

In the number plate example, we select 2 colors out of 3, in the bouquet example we select 2 flowers out of 3. There are 6 ways of doing the former, but only 3 ways for the latter. What is the difference?

**The Idea of Order**

The difference is characterised by this term called ‘order’. In the number plate example, ‘black plate–white letters’ is different from ‘white plate–black letters’; whereas in the bouquet example, ‘rose–lily is the same as lily–rose’.

If in a type of selection ‘AB’ is different from ‘BA’, then order is important. If ‘AB’ is same as ‘BA’ then order is not important. Essentially, if the reason for which the selections are made is the same, then order does not matter. In the bouquet example, the flowers are chosen for the ‘same’ bouquet. But in the number plate example, one color is for the background and the other color is for the letters and numbers. Here the reasons are different.

Mike has three types of flowers with him – roses, lilies and violets. Mike decides he has to create a bouquet that has two flowers. In how many ways can he create this?

RL, LV, RV, RR, LL, VV

What is the difference between selecting two distinct flowers and two flowers?

**The Idea of Repetition**

This difference is based on the idea of repetition. If the same element can be selected again, then we allow ‘repetition’. If we are selecting two flowers, then we can select rose and rose. However, if we are looking for two distinct flowers, then rose–rose is not to be counted.

If in a type of selection ‘AA’ is permitted and should be counted, then repetition is allowed. If ‘AA’ cannot be a legitimate selection, repetition is not allowed.

**Understanding the idea of Repetition**

A teacher has a chocolate, a biscuit and a cold–drink with her. She says that whoever gets a question correct would get one of the three as award.

Ram gets the first question right. How many options does he have of choosing his award? 3

Krish gets the second question right. How many options does he have of choosing his award? 2

John gets the third question right. He has only one option for choosing the award.

Next day, the teacher follows a new awards scheme. She gives Star rating, Circle rating and Square rating to students who get questions right.

Ram gets the first question right. How many options does he have of choosing his award? 3

Krish gets the second question right. How many options does he have of choosing his award? 3

John gets the third question right. How many options does he have of choosing his award? 3

In the first instance, selection is made without repetition. In the second instance, repetition is allowed. If repetition is not allowed, the choice–set shrinks with every selection that is made. If repetition is allowed, then the choice–set remains the same. At every stage, there are the same n objects to choose from.

**Rearrangements**

**Ram, Krishna and John decide to take part in a race. If there is to be no tie, in how many ways can the three positions on the race be determined?**

1st Position | Ram | John | John | Krishna | Krishna |
---|---|---|---|---|---|

2nd Position | John | Ram | Krishna | John | Ram |

3rd Position | Krishna | Krishna | Ram | Ram | John |

For the first slot, there are 3 options, for the second one there are two options, for the third there is only one option. Totally, there are 3 × 2 × 1 = 6 = 3! options.

The number of ways of rearranging r objects is given by r!

**A school plans to paint the three floors of its building. The painter can choose from the colour red, blue, yellow, and orange for each floor in the building. In how many ways can he choose the colours to paint the building?**

The painter can choose from four colors for the first floor.

The painter can choose from four colors for the second floor.

The painter can choose from four colors for the third floor.

Total number of options = 4 × 4 × 4 = 64

**A school plans to paint the three floors in its building. The painter can choose from red, blue, yellow, and orange for each floor in the building. Further he decides that each floor should have a different colour. In how many ways can he choose the colours to paint the building?**

The painter can choose from four colors for the first floor.

The painter can choose from three colors for the second floor.

Total number of options = 4 × 3 × 2 = 24

**The principal of a school, who is an eccentric person, decides that all three floors should have a color that is a mixture of three colours that the painter has selected. In how many ways can the school be painted?**

Now, this is different from the previous question in one key aspect. In the previous question, we select a color for the first floor, one for the second floor and one for the third floor. Here we are going to select 3 colours out of 4 and mix them. Here, a selection of red, blue and green results in an identical outcome as that of a selection of blue, green and red. In other words, order does not matter. The number of ways would be red–blue–yellow, red–blue–orange, red–yellow–orange, and blue–yellow–orange. There are only 4 options totally.

Now, let us see if there is a method to arrive at this answer. Let us look at this by enumerating some options. Look at the following six sequences. These are options that one could have chosen if one were painting each floor with a different colour (as outlined in the previous question).

First | Red | Red | Blue | Blue | Yellow | Yellow |
---|---|---|---|---|---|---|

Second | Blue | Yellow | Red | Yellow | Red | Blue |

Third | Yellow | Blue | Yellow | Red | Blue | Red |

In the case of selecting some set of three colours to get an overall blend, all the 6 options outlined above result in only one end outcome. So, the 4 × 3 × 2 options of the previous question result in (4 x 3 x 2)/6 options for this one. In other words, the number of ways of selecting 3 paints for a blend is (4 x 3 x 2) / ( 3 x 2 x 1) . Why is it 3 × 2 × 1 in the denominator? For every 3 × 2 ×1 outcomes for Question 2, there is only one end outcome when we consider the scenario outlined in Question 3.

What matters in this question is the combination of the colors and not the order. Hence considering the different possible arrangements of a selection is redundant here. And, we know that 3 distinct things can be rearranged in 3! ways; so we need to divide (4 × 3 × 2) by 3! to get the correct answer.

We are going to redefine the questions differently and create a general framework.

**A school has to paint the ‘r’ floors in its building. The painter can choose from n different colours for each floor in the building. In how many ways can he choose the colours to paint the building?**

From n colours, select one for each floor, ‘r’ number of times. This can be done in

From n options, select ‘r’ such that order is important and repetition is allowed. This can be done in

**A school has to paint the ‘r’ floors in its building. The painter can choose from ‘n’ different colours for each floor in the building. Further he decides that each floor should have a different colour. In how many ways can he choose the colours to paint the building?**

From n colours, select one distinct one for each of ‘r’ different floors.

From n options, select ‘r’ three such that order is important and repetition is not allowed. This can be done in n (n – 1) (n – 2) n – 3)….up to r terms. This can be re–written as n! / (n - r)! This term is called nPr.

**The principal of the school, who is an eccentric, decides that all ‘r’ floors should have the same colour and that colour should be a mixture of some ‘r’ of the ‘n’ colours available. In how many ways can the school be painted?**

From n colours, select r and then mix them together to get one blend to be used across the entire school

From n options, select ‘r’ such that order is not important and repetition is not allowed. This can be done in n (n – 1) (n – 2) (n – 3)….up to r terms / (1 × 2 × 3 × …r). This can be re–written as

Also, the number of permutations can be seen as the number of combinations multiplied by the different arrangements possible for each combination. That is, nPr = nCr × r!

Selecting in order = Selecting without order AND Rearranging them

**A football team plays a 5–a–side tournament. In all these tournaments, there should be 1 forward, 1 defender, 2–midfielders, and 1 goal keeper. 1 of the 5 should also be the captain of the team. Ram is the coach of team ‘Samba’ which has a squad of only 5 people. Let us say the 5 people are A, B, C, D and E.**

**In how many ways can Ram select a forward and a defender from this five?**

Number of ways of selecting forward = 5.

Number of ways of selecting defender = 4.

Total number of outcomes = 5 × 4 = 20

The selections are as follows:

BA | CA | DA | EA | |
---|---|---|---|---|

AB | CB | DB | EB | |

AC | BC | DC | EC | |

AD | BD | CD | ED | |

AE | BE | CE | DE |

**In how many ways can Ram select a goal keeper and captain from this 5?**

Number of ways of selecting goal keeper = 5.

Number of ways of selecting captain = 5.

Total number of outcomes = 5 × 5 = 25

The sections are as follows:

AA | BA | CA | DA | EA |
---|---|---|---|---|

AB | BB | CB | DB | EB |

AC | BC | CC | DC | EC |

AD | BD | CD | DD | ED |

AE | BE | CE | DE | EE |

**In how many ways can Ram select 2 midfielders from this 5?**

Number of ways of selecting two players out of

5 = 5C2 = (5 x 4)/2 = 10.

The selections are as follows

AB | ||||
---|---|---|---|---|

AC | BC | |||

AD | BD | CD | ||

AE | BE | CE | DE |

Technically speaking, the three answers are 5P2,

#### Demystifying Permutation & Combination - Rajesh Balasubramanian - CAT 100th Percentile - CAT 2011, 2012 and 2014 (Part 2/7)

**PROBLEMS BASED ON DIGITS OF A NUMBER**

**How many three digit numbers exist? How many of these three–digit numbers comprise only even digits? In how many 3–digit numbers is the hundreds digit greater than the ten’s place digit, which is greater than the units’ place digit?**

Three digit numbers range from 100 to 999. There are totally 900 such numbers. There is a simple framework for handling digits questions.

Let three digit number be ‘abc’.

‘a’ can take values 1 to 9 {as the leading digit cannot be zero}.

‘b’ can take values 0 to 9.

‘c’ can take values 0 to 9.

Totally, there are 9 × 10 × 10= 900 possibilities.

Now, three digit number with even digits

Let the three–digit number be ‘abc’.

‘a’ can take values 2, 4, 6 or 8 {as the leading digit cannot be zero}.

‘b’ can take values 0, 2, 4, 6 or 8.

‘c’ can take values 0, 2, 4, 6 or 8.

4 × 5 × 5 = 100 numbers

**In how many 3–digit numbers is the hundreds digit greater than the ten’s place digit, which is greater than the units’ place digit?**

The digits have to be from 0 to 9. Of these some three distinct digits can be selected in 10C3 ways. For each such selection, exactly one order of the digits will have the digits arranged in the descending order.

So, number of possibilities = 10C3 = 120.

We do not have to worry about the leading digit not being zero, as that possibility is anyway ruled out as a > b > c.

**How many 4–digit numbers exist where all the digits are distinct?**

Let 4–digit number be ‘abcd’.

‘a’ can take values from 1 to 9. 9 possibilities

‘b’ can take values from 0 to 9 except ‘a’. 9 possibilities

‘c’ can take values from 0 to 9 except ‘a’ and ‘b’. 8 possibilities

‘d’ can take values from 0 to 9 except ‘a’, ‘b’ and ‘c’. 7 possibilities

Total number of outcomes = 9 × 9 × 8 × 7

**PROBLEMS BASED ON REARRANGEMENT OF LETTERS OF A WORD**

**In how many ways can we rearrange the letters of the word ‘MALE’?In how many ways can we rearrange the letters of the word ‘ALPHA’?In how many ways can we rearrange the letters of the word ‘LETTERS’?**

Number of ways of arranging letters of the word MALE = 4! = 24. (Think about the number of ways of arranging r distinct things).

Now, ‘ALPHA’ is tricky. If we had 5 distinct letters, the number of rearrangements would be 5!, but here we have two ‘A’s.

For a second, let us create new English alphabet with A1 and A2. Now the word A1LPHA2 can be rearranged in 5! ways. Now, in this 5! listings we would count A1LPHA2 and A2LPHA1, both of which are just ALPHA in regular English. Or, we are effectively double–counting when we count 5!. So, the total number of possibilities = 5!/2.

The formula is actually 5!/2!. . Whenever we have letters repeating we need to make this adjustment.

In how many ways can we rearrange the letters of the word ‘LETTERS’? – 7!/2!2!

**In how many ways can we rearrange letters of the word ‘POTATO’ such that the two O’s appear together?In how many ways can we rearrange letters of the word ‘POTATO’ such that the vowels appear together?**

The two O’s appear together

Let us put these two O’s in a box and call it X.

Now, we are effectively rearranging the letters of the word PTATX. This can be done in 5!/2! ways.

Now, we need to count the ways when the vowels appear together.

Let us put the three vowels together into a box and call it Y.

We are effectively rearranging PTTY. This can be done in 4!/2! ways.

However, in these 4!/2! ways, Y itself can take many forms.

For instance, a word PTTY can be PTTAOO or PTTOAO or PTTOOA.

How many forms can Y take?

Y can take 3!/2! = 3 forms.

So, total number of ways = 4!/2! × 3!/2! = = 12 × 3 = 36 ways

**PROBLEMS BASED ON DICE**

Dice questions have a similar framework to digits question. When a die is thrown thrice, we can take the outcomes to be ‘a’, ‘b’, ‘c’. There are two simple differences vis-a-vis digits questions.

- a, b, c can take only values from 1 to 6. In digits we have to worry about 0, 7, 8 and 9 as well
- There are no constraints regarding the leading die. All throws have the same number of options.

So, in many ways, dice questions are simpler versions of digits questions.

**In how many ways can we roll a die thrice such that all three throws show different numbers?**

Let the throws be ‘a’, ‘b’, ‘c’.

‘a’ can take 6 options – 1 to 6

‘b’ can take 5 options – 1 to 6 except ‘a’

‘c’ can take 4 options – 1 to 6 except ‘a’ and ‘b’

Total number of outcomes = 6 × 5 × 4 = 120

**In how many ways can we roll a die thrice such that at least two throws show the same number?**

We can have either two throws same or all three same.

There are 6 ways in which all three can be same — (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6).

Now, two throws can be same in three different forms a = b, b = c or c = a

a = b –> Number of outcomes = 6 × 1 × 5. ‘a’ can take values from 1 to 6. b should be equal to ‘a’ and c can take 5 values – 1 to 6 except ‘a’

b = c there are 30 possibilities,

and c = a there are 30 possibilities

Total number of options = 6 + 30 + 30 + 30 = 96

**In how many ways can we roll a die twice such that the sum of the numbers on the two throws is an even number less than 8?**

Sum can be 2, 4 or 6

2 can happen in one way: 1 + 1

4 can happen in 3 ways: 1 + 3, 2 + 2, 3 + 1

6 can happen in 5 ways: 1 + 5, 2 + 4, 3 + 3, 4 + 2,5 + 1

Totally, there are 1 + 3 + 5 = 9 ways.

**PROBLEMS BASED ON COIN TOSSES**

**When a coin is tossed three times, how many ways can exactly one head show up?When a coin is tossed coin 5 times, in how many ways can exactly 3 heads show up?When a coin is tossed 5 times, in how many ways can do utmost 3 heads show up?**

Three coins are tossed, options with one head are HTT, THT and TTH. 3 ways

5 coins are tossed, three heads can be obtained as HHHTT, HTHTH, TTHHH, …etc. Obviously this

is far tougher to enumerate.

We can think of this differently. All the versions are nothing but rearrangements of HHHTT. This can be

done in 5!/3!2! ways.

Coins questions are common, so it helps to look at them from another framework also.

Let us assume the outcomes of the 5 coin tosses are written down in 5 slots.

Now, suppose, we select the slots that are heads and list them down.

So, a HHHTT would correspond to 123.

HTHTH would be 135.

TTHHH would be 345.

The list of all possible selections is nothing but the number of ways of selecting 3 slots out of 5. This

can be done in 5C3 ways, or, 10 ways.

Number of ways of getting exactly r heads when n coins are tossed = nCr

In how many ways can we toss a coin 5 times such that there are **utmost** 3 heads?

Utmost 3 heads => Maximum of three heads

Number of ways of having 3 heads = 5C3 = 10

Number of ways of having 2 heads = 5C2 = 10

Number of ways of having 1 head = 5C1 = 5

Number of ways of having 0 heads = 5C0 = 1

Utmost 3 heads = 10 + 10 + 5 + 1 = 26 ways.

**When a coin is tossed 6 times, in how many ways do exactly 4 heads show up? Exactly 2 tails?When a coin is tossed 6 times, what is the total number of outcomes possible?**

Coin tossed 6 times, number of ways of getting 4 heads = 6C4 = 15

Exactly two tails = 6C2 = 15

Every outcome where there are 4 heads corresponds to an outcome where there are 2 tails. So, we are

effectively counting the same set of outcomes in both scenarios.

In other words nCr = nC(n–r)

Total number of outcomes = 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64.

This 64 is also

When a coin is tossed once, there are 2 possible outcomes.

When it is tossed 6 times there will be 2 × 2 × 2 × 2 × 2 × 2 =

Therefore, nC0 + nC1 + nC2 + ……….. + nCn =

{This is also seen in the topic ‘Binomial Theorem’. But never hurts to reiterate!}

**PROBLEMS BASED ON CARD PACKS**

Questions based on card pack are very straightforward. We need to know exactly what lies inside a card pack. Once we know this, everything else falls in place.

A card pack has 52 cards - 26 in red and 26 in black.

There are 4 suits totally, 2 of each colour. Each suit comprises 13 cards - an Ace, numbers 2 to 10, and Jack, Queen and King.

The cards with numbers 2 to 10 are called numbered cards.

Cards with J, Q, K are called Face cards as there is a face printed on them.

**In how many ways can we select 4 cards from a card pack such that all are face cards?**

There are 12 face cards in a pack. Number of ways of selecting 4 out of these = 12C4

**In how many ways can we select 3 cards from a card pack such that none are black numbered cards?**

There are 18 black numbered cards. If we select 3 cards and none of these are black numbered cards, then all of these must be from the remaining 34. Number of ways of selecting 3 cards from 34 is 34C3

**In how many ways can we select 5 cards from a card pack such that we select at least 1 card from each suit?**

We should select 1 card each from 3 of the suits and 2 from the fourth.

The suit that contributes the additional card can be selected in 4C1 ways.

So, total number of outcomes = 4C1 × 13C1 × 13C1 × 13C1 × 13C2

**CIRCULAR ARRANGEMENT**

Let us say there are n people to be seated around a circular table. In how many ways can this happen? If we think about this as n slots, where n things are to fit in, the answer would be n!. But, what we miss out here is the fact that if every object moves one step to the right/left then this would not be a different arrangement. In fact any rotation of one arrangement is not giving us another arrangement. So, how do we think about this?

Let us fix one person’s position. Then, we have the remaining (n–1) persons who can be arranged in (n–1)! ways.

**In how many ways can 6 people be arranged around a circle?In how many ways can 6 people be arranged around a circle if A and B should never sit together?**

Number of ways of arranging n people around a circle = (n –1)!.

So, the number of ways of arranging 6 people around a table = (6 –1)! = 5! = 120.

Now, A and B should never sit together. Let us calculate all the possibilities where A and B do sit together.

Now, A and B can be together called X.

Number of ways of arranging 5 around a circle = 4!.

Now, AB can be sitting such that A is to the left of B or B is to the left of A.

So, total number of options = 4! × 2.

Number of options where A and B do not sit adjacent to each other = 5! – 2 × 4! = 120 – 48 = 72.

**SELECTING ONE OR MORE FROM A SET**

**If there are 4 books and 3 CDs on a table, in how many ways can we select at least one item from the table?**

Each article has 2 options, either you can select it or you can skip it.

Total no of option =

Since we need to select at least least one item, we need to subtract this possibility.

Total number of ways =

**If there are 4 identical copies of a book and 3 identical copies of a CD on a table, in how many ways can we select at least one item from the table?**

We can select either 0, 1, 2, 3 or 4 books. Similarly, we can select 0, 1, 2 or 3 CDs.

So, total number of options = 5 × 4 = 20 ways.

Of these one will include the option of not selecting any of the things.

So, total number of possible outcomes = 4 × 5 – 1 = 19.

Note that here we do not worry about WHICH CD or book we are selecting. Since the CDs and books

are identical, only the number of CDs/books matters

#### Demystifying Permutation & Combination - Rajesh Balasubramanian - CAT 100th Percentile - CAT 2011, 2012 and 2014 (Part 3/7)

**DISTRIBUTION INTO GROUPS**

**In how many ways can we split 8 different objects into groups of 5 and 3?In how many ways can we split this into groups of 4, 3 and 1?In how many ways can we split this into groups of 4, 2, and 2?**

Number of ways of splitting 8 different objects into groups of 5 and 3 = 8C5:

Select 5 objects, the remaining 3 form the second group automatically.

You should also note that it is same as selecting 3 objects from 8, 8C3 x 8C5 = 8C3.

8 different objects into groups of 4, 3 and 1:

First select 4 objects – this can be done in 8C4 ways.

Then select 3 out of the remaining 4 – this can be done in 4C3 ways.

Total number of ways = 8C4 × 4C3 = 8!/4!4! x 4!/3!1! = 8!/4!3!1!

So, it follows that p + q + r objects can be split into groups of p, q and r in (p+q+r)!/p!q!r! ways as long as p, q, r are distinct.

8 different objects into groups of 4, 2 and 2:

Now, we need to treat this differently. From 8 objects, number of ways of selecting 4 is 8C4.

Post this, number of ways of selecting 2 out of 4 would be 4C2.

But 8C4 × 4C2 would overstate the number. In the final 4C2, we calculate the number of ways of selecting 2 objects from 4, with the assumption that the remaining 2 would form the other group.

However, let us say we need to select two out of A, B, C, and D. For every selection of AB in the Group 1 and CD for Group 2, we could have an exact mirror selection CD in Group 1 and AB in Group 2. So, the correct answer should be (8C4 x 4C2)/2!.

Every time we split and allocate into groups of same size, we need to be careful.

**DISTRIBUTION INTO GROUPS – VARIANTS**

**In how many ways can 6 identical toys be placed in 3 identical boxes such that no box is empty?**

This is simple. Since the toys and boxes are identical, we just need to deal with a ways of splitting six into three natural numbers

1 + 2 + 3 = 6

1 + 1 + 4 = 6

2 + 2 + 2 = 6

These are the three ways in which it can be done.

**In how many ways can 6 identical toys be placed in 3 distinct boxes such that no box is empty?**

Again, we are looking to solve for a + b + c = 6. But in this case, a (1, 2, 3) will be counted as a separate from (3, 2, 1) and (2, 1, 3).

(1, 2, 3) can be rearranged in 3! = 6 ways

(1, 1, 4) can be rearranged in 3C1 ways = 3 ways.

We need to select which box has the set of 4 toys. The other two boxes will have 1 each. Alternatively, we are thinking about in how many ways we can rearrange 114. This can be done in 3!2! ways.

(2, 2, 2) can be done in only one way.

So, totally we have 6 + 3 + 1 = 10 ways.

Alternative Method

a + b + c = 6. Now, let us place six sticks in a row

| | | | |

This question now becomes the equivalent of placing two ‘+’ symbols somewhere between these sticks. For instance

| | + | | | + |, This would be the equivalent of 2 + 3 + 1. or, a = 2, b = 3, c = 1.

There are 5 slots between the sticks, out of which one has to select 2 for placing the ‘+’s.

The number of ways of doing this would be 5C2 = 10.

Bear in mind that this kind of calculation counts ordered triplets. (2, 3, 1) and (1, 2, 3) will both be counted as distinct possibilities. This is why we use this method for finding the number of ways of placing 6 identical objects in 3 distinct boxes. So, the above question can also be phrased like this: In how many ways can we have ordered triplets of natural numbers (a, b, c) such that a + b + c = 6. There is another version of this question with ordered triplets of whole numbers. Think about what adjustment needs to be done there.

**In how many ways can 6 distinct toys be placed in 3 identical boxes such that no box is empty?**

First let us think of the distributions.

The boxes can have**1, 2, 3:** This can be done in 6C3 × 3C2 ways. First select 3 out of 6, and then 2 out of the remaining 3. This is nothing but distributing 6 as 3, 2, 1 which can be done in 6!/(2! × 3! ×1!)ways**1, 1, 4:** This can be done in 6C4 ways. Once we select 4 out of 6, the other two go into one box each. Since the boxes are identical, we do not have to worry about selecting anything beyond the first set of 4 toys.**2, 2, 2:** This looks like it could be 6C2 × 4C2 ways. But this will carry some multiple counts. The idea we are using here is simple – select 2 out of 6 and then select 2 out of 4.

When we do this, a selection of AB, and then CD will get counted. This will get accounted as AB, CD, EF. However, we will also be counting a selection of CD, AB, EF, and EF, AB, CD. Since the boxes are identical, all these selections are effectively the same. So, number of ways would be 6C2 x 4C2 / 3!

So, total number of ways of doing this would be 60 + 15 + 15 = 90 ways.

**In how many ways can 6 distinct toys be placed in 3 distinct boxes such that no box is empty?**

Again, let us start with the distributions.

Scenario I: (1, 2, 3): This can be done in 6C3 × 3C2 × 3! ways. First select 3 out of 6, and then 2 out of the remaining 3. After we have done this, the toys can go into the three distinct boxes in 3! ways. 360 ways

Scenario II: 1, 1, 4: This can be done in 6C4 × 3! ways. Once we select 4 out of 6, the other two go get broken up as 1 and 1. Now, we have something akin to ABCD, E and F to be allotted into 3 distinct boxed. This can be done in 3! ways. 90 ways

Scenario III: 2, 2, 2: This should be 6C2 × 4C2 ways. The idea we are using here is simple – select 2 out of 6 for the first box and then select 2 out of 4 for the second box. 90 ways.

Total number of ways = 360 +90 + 90 = 540 ways.

Now, this question can be rephrased wonderfully like this:

How many onto functions can be defined from {a, b, c, d, e, f} to {1, 2, 3}?

You can solve the above question by thinking of all functions from the first set to the second and subtracting the non–onto functions from that. Needless to say, we would get the same answer.

**COUNTING AND NUMBER THEORY**

**How many factors of $2^7 * 11^5 * 7^4$ are perfect squares?**

Any factor of

a < = 7

b < = 5

c < = 4

Any perfect square’s prime factorisation will have all the powers as even numbers. So, a can take values 0, 2, 4, 6; b can take values 0, 2 or 4; and c can take values 0, 2 or 4.

Number of factors that are perfect squares are 4 × 3 × 3 = 36

**All numbers from 1 to 250 (in decimal system) are written in base 7 and base 8 systems. How many of the numbers will have a non–zero units digit in both base 8 and base 7 notations?**

A number when written in base 8, if it ends in 0, should be a multiple of 8. Likewise for base 7. So, effectively this question becomes — How many natural numbers exist less than 251 that are multiples of neither 7 nor 8.

Let us first find out numbers that are multiples of either 7 or 8.

Multiples of 7 — 7, 14, 21, 28,.......245... 35 numbers in this list.

Multiples of 8 — 8, 16, 24, 32,.......248... 31 numbers in this list.

Some numbers will be multiples of 7 and 8.

Multiples of 56 — 56, 112, 168, 224… 4 numbers in this list

Number of numbers that are multiples of 7 or 8 = 35 + 31 – 4 = 62

Number of numbers that are multiples of neither 7 nor 8 = 250 – 62 = 188

**How many natural numbers less than 10000 exist such that sum of their digits is 6?**

We are considering all numbers up to 9999.

All numbers of the form ABCD such that a, b, c, d can take values from 0 to 9.

a + b + c + d = 6 where a, b, c, d are all whole numbers.

Now, a, b, c, d can take value 0. Let us simplify this as p = a + 1, q = b + 1, r = c + 1, s = d + 1.

Now, p + q + r + s = 10. p, q, r, s cannot be zero.

Number of solutions for the above equation is 9C3.

Bear in mind that p, q, r, s can all never be greater than 10. In this case, as the total adds up to 10, we have little to worry about. If the sum were greater than 10, it could become far more complex.

**How many numbers are factors of $24^{20}$ but not of $24^{15}$ ?**

Number of factors of this number = 61 × 21 = 1281

Number of factors = 46 × 16 = 736

Factors of

So, the number of numbers that are factors of

61 × 21 – 46 × 16

= 1281 – 736

= 545.

**How many natural numbers less than $10^4$ exist that are perfect squares but not perfect cubes?**

Number of perfect squares less than 10000 = 99.

10000 =

So, there are 99 perfect squares less than 10000?

From these some numbers that are also perfect cubes have to be eliminated.

So, we are looking for numbers that are perfect squares and perfect cubes. Or, we are looking for powers of 6.

So, there are only 4 powers of 6.

So, out of 99, we need to subtract 4 possibilities. Or, there are 95 different natural numbers that will be perfect squares but not perfect cubes

Will solve some problems!

**In how many ways can the letters of the word ‘MALAYALAM’ be rearranged? In how many of the words would the A’s appear together? In how many of the words are the consonants together?**

Rearrangements of MALAYALAM = 9!/4!2!2!

Rearrangements where the A’s appear together:

Put 4 A’s together into X and rearrange MLYLMX. This can be rearranged in 6!/2!2!

Rearrangements where the consonants appear together: Put consonants together into Z. The word will be AAAAZ. This can be rearranged in 5!/4! ways.

Now, Z is MMLLY. Z can take 5!/2!2! ways.

So, total number of arrangements =5!/4! × 5!/2!2!

**In Octaworld, everything is written in base 8 form. Ram’s 1050 page tome written in the real world is translated to Octa–speak and reprinted in Octa–world. In the reprint, each page number is written in base 8 form. How many times will the digit 4 be printed on the page numbers?**

So, we need to see how many times the digit 4 gets printed from 1 to 2032 in base 8.

Let us consider a number of the form

When c = 4, a can take all values from 0 to 7 and b can take all values from 0 to 7. So, 4 gets printed at c 64 times.

When b = 4, a can take all values from 0 to 7 and c can take all values from 0 to 7. So, 4 gets printed at b 64 times.

When a = 4, b can take all values from 0 to 7 and c can take all values from 0 to 7. So, 4 gets printed at a 64 times.

So, totally 4 gets printed 64 × 3 = 192 times from

From

So, up to

We need to account for all base 8 numbers from

In these 26 numbers, the digit 4 gets printed thrice

So, the digit 4 gets printed 384 + 3 = 387 times.

**When a die is thrown twice, in how many outcomes will the product of the two throws be 12?**

12 can be formed from 3 × 4 or 2 × 6 {1 × 12 is not possible with die}.

This can happen in 2 + 2 = 4 ways.

**How many words exist that have exactly 5 distinct consonants and 2 vowels?**

Scenario I: 5 distinct consonants and 2 distinct vowels – Number of words = 21C5 × 5C2 × 7!

Scenario II: 5 distinct consonants and 1 vowel appearing twice – Number of words = 21C5 × 5C1 × 7!/2!

**When a coin is tossed 6 times, in how many outcomes will there be more heads than tails?**

We should have more heads than tails => There should be 4 heads or 5 heads or all 6 heads.

Number of ways = 6C4 + 6C5 + 6C6

= 15 + 6 + 1

= 22

**In how many ways can we pick 4 cards from a card pack such that there are no Aces selected and there are more face cards than numbered cards?**

Scenario I: 3 face cards and 1 numbered card: 12C3 × 36C1

Scenario II: 4 face cards and 0 numbered cards: 12C4

Therefore, total number of ways is 12C3 × 36C1 + 12C4

**On a table, there are 4 identical copies of a book and 3 CDs. In how many ways can we pick at least one book and at least one CD from the table?**

There are 4 identical copies of a book. One can pick either 0, 1, 2, 3, or all 4 of these – 5 different options. We need to pick at least one book. So, we have only 4 options – 1, 2, 3, or 4 books being picked

There are 3 different CDs. Each CD can be either picked or not picked. So, total number of options =

Of these there is one option where no CD is picked. We need to exclude that option. So, number of possibilities =

Total number of outcomes = 4 × (

= 4 × 7

= 28

**What is sum of all 4-digit numbers formed by rearranging the digits of the number 2235?**

Number of rearrangements of 2235 = 4!/2! = 12.

So, we need to add these 12 numbers.

Let us consider the units’ digit of these 12 numbers.

The units digit will be the one of the digits 2, 3, or 5. If the last digit were 3, the first 3 digits should be some rearrangement of 2, 2, and 5. So, there are 3!/2! Such numbers. Or, 3 such numbers.

Similarly there are three numbers with 5 as the units digit.

If the last digit were 2, the first 3 digits should be some rearrangement of 2, 3, and 5. So, there are 3! such numbers, or, 6 such numbers.

So, the units digit will be 2 for six numbers, 3 for three numbers, and 5 for three numbers. Sum of all these unit digits will be 2 × 6 + 3 × 3 + 5 × 3 = 12 + 9 + 15 = 36.

Sum of all the tens digits will be 36. Sum of all the digits in the hundreds’ place will be 36.

Sum of all the digits in the thousands’ place is 36.

So, sum of all the 4–digit numbers will be 36 × 1111 = 39996.

**When a die is thrown twice, in how many ways can we have the sum of numbers to be less than 8?**

Sum of the numbers seen in the two throws can be 2, 3, 4, 5, 6 or 7.

Sum of the digits = 2: This can only be 1 + 1. One way

Sum of the digits = 3: This can be 1 + 2 or 2 + 1. 2 ways

Sum of the digits = 4: 1 + 3, 3 + 1, 2 + 2. 3 ways

Sum of the digits = 5: 1 + 4, 4 + 1, 3 + 2, 2 + 3. 4 ways

Sum of the digits = 6: 1 + 5, 5 + 1, 2 + 4, 4 + 2, 3 + 3. 5 ways

Sum of the digits = 7: 1 + 6, 6 + 1, 2 + 5, 5 + 2, 3 + 4, 4 + 3. 6 ways

Sum of the numbers in the two throws can be less than 8 in 1 + 2 + 3 + 4 + 5 + 6 = 21 ways.

We notice a very simple pattern here. Try the sum of the numbers all the way to 12 and see the rest of the pattern also.

**Set P has elements {1, 2, 3..…10}. How many non–empty subsets of P have the product of their elements as not a multiple of 3?**

Total number of subsets =

For choosing any subset, each element can either be part of the subset or not part of the subset. So, for each element, there are two options. So, with 10 elements in the set, we can create

Subsets whose product is not a multiple of 3 = Subsets of the set {1, 2, 4, 5, 7, 8, 10} =

**A, B, C, D, E are doctors, P, Q, R, S, are engineers. In how many ways can we select a committee of 5 that has more engineers than doctors?**

Two scenarios are possible.

3 engineers and 2 doctors: 4C3 × 5C2 = 4 × 10 = 40

4 engineers and 1 doctor : 4C4 × 5C1 = 1 × 5 = 5

Total number of possibilities = 40 + 5 = 45

**From a card pack of 52, in how many ways can we pick a sequence of 4 cards such that they are in order and from different suits? Consider Ace to be the card following King in each suit. So, Ace can be taken to precede ‘2’ and succeed ‘King’. So, JQKA would be a sequence, so would be A234. However, QKA2 is not a sequence.**

4 cards in order can be A234, 2345, ….JQKA. 11 different possibilities

For a given set of four cards, say 2345, they can be from 4 different suits in 4! ways.

So, total number of possibilities = 11 × 4! = 264.

#### Know Your Numbers

** “We all use math every day; to predict weather, to tell time, to handle money. Math is more than formulas or equations; it's logic, its rationality, it’s using our mind to solve the biggest mysteries we know” – Numb3rs**

We started dealing with numbers early in our childhood, but still this area poses a major challenge to majority of us preparing for aptitude tests. We either overlook this topic presuming it to be simple or are intimidated by the concepts related with it. But the truth is that anyone who intends to prepare seriously for an aptitude test is left with no other option but to study number systems top to bottom. Reasons being,

- A big share of questions in Quant comes from Number systems. Once you are comfortable with Number system concepts, the so called herculean task of clearing Quant cutoff shall become much easier.
- Mastery in Number system concepts will give you an extra edge in two very important and calculation intensive sections - Quant and DI.
- Your comfort level with numbers will reflect in your solving speed, saves you time and hence boost your overall score.
- The Concepts handled in Number systems are well known to us and all that is required is a brush up.
- Number system is a major error-prone area. Mistakes not just take away our hard earned marks but also waste precious time.

Let’s make peace with numbers and as the adage goes "A friend in need is a friend indeed”. This friendship will come in very handy during our exams!!

**Natural numbers**

{1, 2, 3, 4…}

Natural numbers have 2 main purposes**counting** (1, 2, 3, etc.)**ordering** (1st, 2nd, 3rd, etc.)

**Whole numbers**

{0, 1, 2, 3…}

Why Zero is not a natural number ?

**Integers**

{…-2, -1, 0, 1, 2…}

Integer is the Latin for "Whole". It denotes positive, zero and negative.

**Decimals**

**There are 3 types of decimal numbers**

- Terminating decimal - has a finite decimal value. (e.g. 10.243 )
- Infinite recurring decimals - will not terminate and has a recurring pattern in it. (e.g.: 4.333333333333333…. , where recurring part is 3)
- Infinite non recurring decimals - will not terminate and has no recurring pattern in it (e.g.: 3.453298098736253322….)

**Fractions**

**A common fraction** (also known as a vulgar fraction or simple fraction) is a rational number written as a/b where the integers a & b are called the numerator and the denominator, respectively.

**Proper fraction**: numerator is smaller than the denominator

absolute value of proper fraction is less than one - fraction is between -1 and 1

**Improper fraction** (also called Top-Heavy fraction) :

numerator is larger than or equal to denominator

absolute value of improper fraction is greater than or equal to 1

**Mixed fraction**: sum of a non-zero integer and a proper fraction

To convert a mixed number into an improper fraction, multiply the whole number by the denominator and add it to the numerator. This becomes the numerator of the improper fraction; the denominator of the new fraction is the same as the original denominator.

**Even numbers**

Any integer that can be represented in the form 2n, where n is an integer

**Odd numbers**

Any integer that can be represented as 2n + 1 or 2n – 1, where n is an integer

Note that even and odd numbers are always integers.

**Positive/Negative**

Positive: Numbers greater than zero

Negative: numbers less than zero

**Real numbers**

Either rational (5, -2, 4/3 …) or irrational (√2, ∏ …)

Either algebraic or transcendental

Either positive, negative, or zero

Every real number can be represented on the number line.

**Rational numbers**

**Numbers which can be represented in the form of p/q, q # 0**

2, -2, 2.2, 2/3, 4.333333… (39/9) are all rational numbers.

Any infinite decimal values with recurring pattern can be represented in p/q form.

Consider 0.33333333… This can be converted to p/q form as 3/9, which is the recurring part divided by n nines where n is the number of digits in the recurring part (here 1).

We can write, 0.464646… = 46/99; 2.383383383… = 2 + 383/999

Now what if the number is 0.34583838383… where recurring pattern (83) starts after some digits and such numbers are called impure recurring decimals.Here method to get the p/q form is to subtract the non recurring part (345) from the number formed by non recurring digits and the recurring pattern written once which will give us the numerator (p). For denominator (q) write as many nines as the number of digits in the recurring part (here 83 is the recurring part, so 99) followed by as many zeros as the number of digits in the non recurring part (here 345, hence three zeros). We get denominator as 99000.

0.34583838383… = (34238 – 345) / 99000 = 34238/99000

**Irrational numbers**

**Any number that cannot be written in p/q form.**

Infinite non recurring decimal values where there is no end and no recurring pattern are irrational numbers. Examples are √2, √3, √5, ∏ etc…

**Prime numbers**

**A natural number greater than unity is a prime number if it does not have a factor except for itself and unity.**

E.g.: 2,3,13,29,53,79 etc…

A simple method of verifying the primality of a given number n is known as trial division. It consists of checking whether n is divisible by any prime number which is less than √n. If not, then n is a prime number.

For 29, √29 is between 5 and 6. Prime numbers less than 6 are 2, 3 and 5. Number 29 is not divisible by any of these numbers making 29 a prime number.

Prime numbers of the form 2^{p}−1 are Mersenne primes (P – Prime number)

A prime number p is a Sophie Germain prime if (2p+1) is also a prime.

Prime numbers whose difference is 2 are called twin primes. e.g.: (3, 5), (11,13)

Prime numbers less than 1000 (168 prime numbers) are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997.

**Composite**** numbers**

A composite number is a positive integer that has at least one positive divisor other than one or itself.

**One is neither prime nor composite.**

**Every composite number can be written as the product of two or more primes.**

E.g. 100 = 2^{2}^{}* 5^{2}

A composite number with two prime factors is a** semi prime**.

A composite number with three distinct prime factors is a **sphenic number.**

If none of its prime factors are repeated, it is called **square free**.

**Powerful number**

A powerful number is a positive integer m such that for every prime number p dividing m, p^{2} also divides m.

A powerful number can be written as the product of a square and a cube, that is, m = a^{2}b^{3 }(a, b are positive integers).

E.g. 1, 4, 8, 9, 16, 25, 27…

**ZERO**

Zero is neither positive nor negative.

Indian mathematician Brahmagupta's ‘**Brahmasphu****ṭ****a****siddhanta**’ is the first book that mentions zero as a number; hence Brahmagupta is considered the first to formulate the concept of zero.

Zero fits the definition of even number: it is an integer multiple of 2, namely 0 × 2.

Zero is not a natural number as one normally does not start counting with zero. Yet zero does represent a counting concept: the absence of any objects in a set. To resolve this issue, some mathematicians define the natural numbers as the positive integers.

Zero is neither a prime number nor a composite number.

**π (PI)**

The number π is a mathematical constant that is the ratio of a circle's circumference to its diameter, and is approximately equal to 3.14159.

π is an irrational number, which means that it cannot be expressed exactly as a ratio of two integers (22/7 is commonly used to approximate π)

π is a transcendental number – a number that is not the root of any nonzero polynomial having rational coefficients.

The digits of π have no apparent pattern

**e**** (****Napier's constant)**

The number e is approximately equal to 2.71828

e is the base of the natural logarithm.

e is irrational

e is transcendental

**Perfect Squares**

An integer that is the square of an integer.

Squares are always non negative.

A positive integer that has no perfect square divisors except 1 is called square-free.

**Square root**

y is a square root of number x if y^{2}= x

Every positive number x = y^{2}has two square roots, +y and -y.

Square roots of positive whole numbers that are not perfect squares are always irrational numbers

**Perfect numbers**

If the sum of all the positive factors of a number (known as aliquot sum) is equal to the number itself, then the number is a perfect number.

Example: 6 (factors are 1, 2, 3 and 6). 1+2+3 = 6

28 is a perfect number as 1 + 2 + 4 + 7 + 14 = 28

It is unknown whether there are any odd perfect numbers.

**A Square number cannot be a perfect number.**

The only even perfect number of the form x>^{3} + 1 is 28. 28 is also the only even perfect number that is a sum of two positive integral cubes (27 + 1).

The reciprocals of the divisors of a perfect number N must add up to 2

(for 6, 1/6 + 1/3 + 1/2 + 1 = 2).

Euclid proved that 2^{p}^{−}^{1}(2^{p}−1) is an even perfect number whenever 2^{p}−1 is prime

for p = 2: 2^{1}(2^{2}−1) = 6

for p = 3: 2^{2}(2^{3}−1) = 28

for p = 5: 2^{4}(2^{5}−1) = 496

for p = 7: 2^{6}(2^{7}−1) = 8128.

**Co Primes ****(or relatively primes)**

Two numbers are said to be co primes if the only common factor between then is one.

E.g.: (5, 4), (3, 10)

Two consecutive integers are always co primes.

**Complex numbers**

a number which can be put in the form of a + **i** b where a and b are real numbers and **i **is called the imaginary unit

**i**= √-1. ( **i**^{2}= -1 )

**Triangular numbers**

Sum of n consecutive integers starting with 1.

E.g.: 1 + 2 = 3

1 + 2 + 3 = 6

1 + 2 + 3 + 4 = 10

Why it is called Triangular numbers? Because using ‘triangular number’ of objects we can form an equilateral triangle. Check it out!

All even perfect numbers (till we get our odd one) is a triangular number.

**Armstrong number****(narcissistic number)**

Is an integer such that the sum of their own digits to the power of the number of digits is equal to the number itself. 371 = 3³+ 7³+ 1³, 153 = 1³+ 5³+ 3³

**Kaprekar number**

Consider n digit number K. Square the number and if the sum first n digits from rightand the n ( or n-1) digits from left of K^{2 }is equal to K it is a Kaprekar number.

For example: K = 9, n =1. K^{2 }= 81, now 8 + 1 = 9 = K;

K = 703, n =3. K^{2}= 494209, 494 + 209 = 703 = K.

**Bell number**

The number of ways a set of n elements can be partitioned into nonempty subsets is called a Bell number.

For example, there are five ways the numbers {1,2,3} can be partitioned into nonempty sets: {{1},{2},{3}}, {{1,2},{3}}, {{1,3},{2}}, {{1},{2,3}}, and {{1,2,3}}, so Bellnumber(3) = 5.

**Palindromes**

A palindrome number is a number that is the same when written forwards or backwards. Example: 44, 121, 8778

**Automorphic numbers**

A number k such that n.k^{2} has its last digit(s) equal to k is called n-automorphic.

For example, 1·5^{2}=25 and 1·6^{2}=36, so 5 and 6 are 1-automorphic.

2·8^{2}=128 and 2·88^{2}=15488, so 8 and 88 are 2-automorphic.

**Ramanujan number ( 1729 )**

It is the smallest number expressible as the sum of two positive cubes in two different ways.

1729 = 1^{3}+ 12^{3}= 9^{3}+ 10^{3}

**Smith number**

A Smith number is a composite number the sum of whose digits is the sum of the digits of its prime factors.

The primes are excluded since they trivially satisfy this condition.

Example of a Smith number is the **Beast number (666)**

666=2·3·3·37

6+6+6=2+3+3+ (3+7) =18.

**Amicable Pair**

An amicable pair (m, n) consists of two integers m, n for which the sum of proper divisors (the divisors excluding the number itself) of one number equals the other.

Example: (220,284)

**Friedman number**

A Friedman number is a positive integer which can be written in some non-trivial way using its own digits, together with the symbols + - x / ^ ( ) and concatenation. For example, 25 = 5^{2}

Sharing some Friedman numbers, can you find out how the digits of below numbers can be used to generate the number itself with the operations allowed.

121, 125, 127, 128, 153, 216, 289, 343, 347, 625, 688, 736, 1022, 1024

**Some useful properties of numbers**

Sum of first n natural numbers = **n * (n + 1) / 2**

Sum of first n odd natural numbers is equal to **n**^{2}

Sum of first n even natural numbers is equal to **n (n + 1)**

Sum of squares of first n natural numbers = **n (n + 1) (2n + 1) / 6**

Sum of cubes of the first n natural nos. = **[n (n + 1) / 2] **^{2}

even ± even = even

even ± odd = odd

odd ± odd = even

even × even = even

even × odd = even

odd × odd = odd

**Goldbach's conjecture** Every even integer greater than 2 can be represented as a sum of two prime numbers ; 12 = 5 + 7

Any positive integer can be written as the sum of four or fewer perfect squares.

14 = 4 + 1 + 9 **(Lagrange's four-square theorem)**

Any prime number of the form (4n + 1) can be written as sum of two squares.

17 = 16 + 1, 41 = 25 + 16

A perfect square can end only with digit 1, 4, 6, 9, 5 or even number of zeros.

Adding the digits of perfect square will always yield 1, 9, 4 or 7. A number is not a perfect square if its digit sum does not evaluate to one of these. Number may or may not be a perfect square if its digit sum is 1, 9, 4 or 7.

Digit sum (1936) = digit sum (1 + 9 + 3 + 6 = 19) = digit sum (1 + 9 = 10) = 1

All prime numbers can be represented as (6n-1) or (6n+1). So if we cannot represent a number in (6n+1) or (6n-1) form we can say it is not a prime. The reverse may not be true, i.e. all numbers that can be represented as (6n-1) or (6n+1) need not be a prime.

Any odd number is a difference of two consecutive squares

(k + 1)^{2}= K^{2}+ 2k +1^{2}, so (k + 1)^{2}- k^{2}= 2k + 1 (an odd number)

Any multiple of 4 is a difference of the squares of two numbers that differ by two

(k + 2)^{2}– k^{2}= 4k + 4 = 4 (k +1)

3^{0}, 3^{1}… 3^{n} can produce all integers from 1 to 1 + 3^{1} + ... + 3^{n}

**Let’s play …**

**1.** Three times the first of three consecutive odd numbers is 3 more than twice the third. What is the third integer?

**2.** If 8 + 12 = 2, 7 + 14 = 3 then 10 + 18 =?

**3.** If 38 pieces of clothing (including shirts, trousers and ties) are sold where at least 11 items in each category is sold. What will be the number of ties must have sold if more shirts are sold than trousers and more trousers than ties?

**4.** A 2 digit number exceeds by 19 the sum of the squares of the digits and by 44 the double product of its digits. Find the number.

**5. **95, 52, 43, 148, 62, 86, 196, __, 121.

**6.** 24:15:: 63: __

**7.** 0, 6, 24, 60, 120, ___

**8.** 6, 14, 36, 98, 276, ___

**9.** 5 + 3 + 2 = 151022

9 + 2 + 4 = 183652

8 + 6 + 3 = 482466

5 + 4 + 5 = 202541

7 + 2 + 5 =?

**10.** Try making 1000 with eight 8s and only addition is allowed.

**11.** Create an eight digit number using the digits 4, 4, 3, 3, 2, 2, 1 and 1. So that, 1s are separated by 1 digit, 2s are separated by 2 digits, 3s are separated by 3 digits and 4s are separated by 4 digits.

**12.** It takes 1629 digits to number the pages of a book. How many pages does the book have?

**13.** What is the minimum number of weights required to weigh from 1kg to 40 kg (integer values only)

**14.** Prove that 121 is a Friedman number (Obtain the number 121 from the digits of the number (1, 2 and 1) and mathematical operations. Each digit can be used only once)

**15.** Use exactly four 4's to form every integer from 0 to 10, using only the operators +, -, *, /, () (brackets) x^{2} (square) Factorial (4! = 24) and decimal point.

**Solutions**

**1.** 15.

Take the numbers as 2n+1, 2n+3 and 2n+5

3* (2n+1) = 2 * (2n+5) + 3

Solve and we get n = 5, so third integer is 2*5 + 5 = 15.

**2.** Sometimes we may not see the direct answer in the options. Answer will be again manipulated through some method. Here what we need is the sum of the digits in the answer, 2 + 8 = 10

**3.** 11 ties are sold.

Consider x Shirts, y trousers and z ties are sold

x + y + z = 38; x, y, z >=11; x > y > z

only possible answer is x = 14, y = 13 and z = 11.

**4.** 72.

Let the digits be a and b, number can be written as 10a + b

10a + b = a^{2}+ b^{2 }+ 19 --- (1)

10a + b = 2ab + 44 --- (2)

(1) – (2), a - b = 5, a = b+5, replace ‘a’ in (1), we get b=2, a=2+5=7

**5. **75. Series can be obtained as 95 - 52 = 43, 148 - 62 = 86,196 - 75 = 121

**6.** 24:15 = 5^{2}-1: 4^{2}-1; 63 = 8^{2}-1, hence the missing number is 7^{2}-1 = 48

**7.** 210. Series can be obtained as nth term = (n-1) n (n+1)

**8.** 794. Series can be obtained as nth term = 1^{n }+ 2^{n }+ 3^{n}

**9.** 143547 (7 * 2, 7 * 5, 7 * 2 + 7 * 5 -2)

**10.** 888 + 88 + 8 + 8 + 8 = 1000

**11.** 41312432; 23421314

**12.** Page1 to Page9 we need 9 digits. Page10 to Page99 we need 180 digits. We are left with 1629 – 180 – 8 = 1440 which can gives us 480 (1440/3) pages with 3 digit page numbers (100-579), we have total 579 pages.

**13. ** 4 weights, 1 kg, 3 kg, 9 kg, 27kg. We can measure all values from 1 and 40. e.g.: to measure 5 kg have 9kg weight at one pan and 3 kg + 1 kg weight on the other.

**14.** 121 = 11^{2}

**15. ** 0 = 44 – 44; 1 = 44 / 44 ; 2 = 4/4 + 4/4 ; 3 = (4 + 4 + 4) / 4; 4 = 4 * (4 - 4) + 4 ; 5 = (4 * 4 + 4) / 4 ; 6 = 4 * .4 + 4.4 ; 7 = (44 / 4) – 4 ; 8 = 4 + 4.4 - .4 ; 9 = 4/4 + 4 + 4 ; 10 = 44 / 4.4

I am sure you know more number types and number properties.. do share :-)

**Happy Learning!** :)

#### Play With Numbers

Calculation speed is not a chapter to learn but a skill to achieve. One among the most hyped element in aptitude test preparation is about achieving that impeccable calculation speed where we need not even touch a pen to get our answer. Sounds great! But at exam room, chances are more that we need our pen! Aspirants get overwhelmed with the idea of calculation speed and start learning methods that can help in quick solving.

Some of these techniques are really helpful, but most of them are helpful only when we are practicing it for a period of time and got habituated with the same.Here the trap is if we decide to use some method just because it is considered as a “Shortcut”, we probably end up rechecking answers with our natural methods! Shortcuts don’t always save time. Methods we know for more than a decade are more likely to help us under exam pressure. Still, some techniques are easy to learn and handy to use. Some such concepts are shared here. Keep practicing and tweaking these methods so that they becomes the natural way of calculation. Two major sections in aptitude exams (Quant & DI) are calculation intensive. Even if we save 15 seconds from a question through smart approach we should be able to attend more questions or improve our accuracy. Considering each mark scored (or lost) has a huge impact on the result, you do the math. Literally! :-)

The main differentiator between people who consistently score good marks in quant and others who doesn’t is the way they approach their basic math like multiplication and division. Those who are comfortable with quant save a lot of time in calculation and they don’t mess up with their accuracy. If you think you are not getting enough time in quant section, call your math loving friends for a coffee and "steal" their calculation methods :)

Most of us require lesser time for addition and subtraction when compared to multiplication and division… when life teach us to use our strength over weakness, why not here!

Multiplication and division can be represented as addition and subtraction operations. 39 * 46 = 1794 says that the number 39 when added 46 times yields 1794 or when 46 is subtracted 39 times from 1794, result is zero. Similarly when we say 819/13 = 63 in math, English will say 63 should be added 13 times to get 819 or 13 should be subtracted 63 times from 819 to reach zero.

We are assuming you have good speed and accuracy in addition/subtraction of 2 digits and 3 digits numbers. If you feel a little bit practice will not hurt then try using the time you spent sitting in a bus or auto… start looking at the number plates around and do your addition/subtraction practice sessions, where else you will get so many numbers flashing before you! Not convinced...? Numbers which are the smallest that can be expressed as the sum of two cubes in n distinct ways have been dubbed as "**taxicab numbers**"!

Here goes the famous anecdote of the British mathematician G. H. Hardy regarding a visit to the hospital to see our great mathematician Srinivasa Ramanujan.

**“I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways." **

**The quotation is sometimes expressed using the term "positive cubes", since allowing negative perfect cubes (the cube of a negative integer) gives the smallest solution as 91.****This ambiguity is eliminated by the term "positive cubes".**** 91 = 6**^{3}**+ (-5)**^{3}**= 4**^{3}**+ 3**^{3}

As a first step, learn squares and square roots of natural numbers till 30 and also be comfortable with some common fractions …

**Knowing square roots also will help to save time in many scenarios.**

**Some useful fractional values are given below**

**Calculation of squares of numbers ending with 5**

If the number is of the form X5 then (X5)^{2}= X * (X + 1) and append 25 at the end.

65^{2}= 6 * (6+1) and append 25 at the end = 4225

(135)^{2}= 13 * (13+1) and append 25 at the end = 18225

This comes handy in scenarios like 85 * 87

= 85 * 85 + 85 * 2 = 7225 + 170 = 7395

This method is actually derived from another useful Vedic math concept. It gives an easy way to find the product of two numbers if their unit digits add up to 10 and rest of the digits are same.

Say, 72 * 78

Units digits adds up to 10 (2 + 8 ) and other digits are same (7).

Here product can be found out as 7 * (7 + 1) (2 * 8 ) = 5616.

Similarly 144 * 146 = 14 * 15 4 * 6 = 21024

Many more neat tricks are there to easily find the square of numbers…

**To calculate squares of two digit numbers**

Let the number be xy then (xy)^{2 }= [x^{2}][y^{2}] + 20 * x * y

Take 58, (58)^{2 }= [5^{2}][8^{2}] + 20 * 5 * 8 = 2564 + 800 = 3364

Take 81, (81)^{2 }= [8^{2}][1^{2}] + 20 * 8 * 1 = 6401+ 160 = 6561 ( note: we wrote 1^{2 }as 01 ( 2 digits) )

**To calculate squares for numbers closer to 100**

Take 94, 94 is 6 less than 100, so 94^{2 }= [94 – 6][6^{2}] = 8836 (note: we **subtracted** the difference as the given number is **lesser** than 100)

Take 108, 108 is 8 more than 100, so 108^{2 }= [108 + 8][8^{2}] = 11664 (note: we **added** the difference as the given number is **greater** than 100)

Take 119, 119 is 19 more than 100, so 119^{2} = [119 + 19] [19^{2}] = [138][361] as the second part is more than 2 digit take the most significant number (here 3) and add it to the first part (138 ).

So 119^{2 }= [138 + 3][61] = 14161

Take 125, 125 is 25 more than 100, so 125^{2} = [125 + 25]... we dont need to do that... we have a better trick, right? 125^{2} = [12 * 13][25] = 15625 **:)**

**To calculate squares of numbers closer to 50**

58 = 50 + 8

58^{2} = [25 + 8] [8^{2}] = 3364 [Note: **Add** the difference to 25 to get the first 2 digits as the given number is **greater** than 50]

63 = 50 + 13

63^{2 }= [25 + 13] [13^{2}] = [38] [169], as the second part is more than 2 digit take the most significant number (here 1) and add it to the first part (38 ).

63^{2 }= [38 + 1][69] = 3969

51 = 50 + 1

51^{2 }= [25 + 1][1^{2}] = 2601 [Note: we wrote 01 as we need 2 digits in each bracket]

47 = 50 – 3

47^{2 }= [25 – 3][3^{2}] = 2209 [Note: **Subtract** the difference to 25 to get the first 2 digits as the given number is **lesser** than 50]

33 = 50 – 17

33^{2 }= [25 – 17][17^{2}] = [8][289], as the second part is more than 2 digit take the most significant number (here 2) and add it to the first part (8 ).

33^{2 }= [8+2][89] = 1089

So now we know how to find squares faster. If just knowing was enough, life would have been much easier… right? We should be able to apply our strengths to our advantage in solving a situation… our confidence in calculating squares can help us in some quick multiplication… How?

We all know the basic formula** (a + b) (a - b) = a ^{2 }– b^{2}**

87 * 93 = ( 90 – 3 ) * (90 + 3) = 90^{2 }– 3^{2} = 8091

similarly, 43 * 47 = 45^{2 }– 4 = 2025 – 4 = 2021

Ok.. Now solve 35 * 45 in 5 seconds :)

**Multiplication**

9 * 12 = ____ Ok, you got 108; question is how you did it?

Most of you might have done it using the logic 9 * 10 + 9 * 2 = 108.

27 * 23 = ___

27 * 20 + 27 * 3 = 540 + 81 = 621

634 * 126 = ___

634 * 126 = 634 * 100 + 634 * 26 = 63400 + 634 * 20 + 634 * 6

= 63400 + 12680 + 3804 = 79884

To find 634 * 6 we used 634 * 5 + 634, an easy way to multiply by 5 is multiply by 10, i.e. put a zero at the end, and divide by 2; 636 * 6 = 3170 + 634 = 3804)

**Try solving the below questions using the above method.**

1. 85 * 11

2. 69 * 71

3. 45 * 43

4. 112 * 120

5. 85^{2}– 65^{2}

6. 814 * 629

**Solutions**

1. 850 + 85 = 935

2. (70-1) * (70+1) = 4900 – 1 = 4899 (a^{2}- b^{2}= (a + b) (a - b))

or 71 * 69 = 71 * 70 - 71 = 70 * 70 + 70 - 71 = 4900 - 1 = 4899

3. 45 * (45 - 2) = 45^{2}– 2 * 45 = 2025 – 90 = 1935

4. 120 (100 + 12) = 12000 + 1440 = 13440

5. 7225 – 4225 = 3000

6. (800+14) * (630-1) = 800 * 630 * 14*630 - 800 - 14,

800 * 630 = 800 * (600+30) = 480000 + 24000 = 504000

14 * 630 = 10 * 630 + 4 * 630 = 6300 + 2520 = 8820

Our answer is 504000 + 8820 – 800 – 14 = 512006

**Division**

Easiest way is to represent numerator as sum/ difference of terms related to denominator.

183 / 16 = ___

183 = 160 (16 * 10) + 16 (16 * 1) + 8 (16 * 0.5) – 1

183 / 16 = 10 + 1 + 0.5 - x (some small value) ≈ little less than 11.5 (approx)

3840 / 23 = ____

3840 = 2300 + 1150 + 230 + 115 + 45

3840 / 23 = 100 + 50 + 10 + 5 + (little less than 2) ≈ little less than 167

There will be a lot of calculations based on fractions waiting for you, mostly in data interpretation section. It’s very important to have a high comfort level with such numbers. Keep below fractions handy.

**Let’s play!**

Which is greater, 10/9 or 7/6?

10/9 = 10 * 0.111 = 1.11

7/6 = 7 * 1/2 * 1/3 = 7 * 0.5 * 0.333 = 3.5 * 0.333 = 1.1655

7/6 > 10/9

Another method is 10 = 9 + 0.9 + 0.09 + …

10/9 = 1 + 0.1 + 0.01 + … = 1.111

7 = 6 + 0.6 + 0.36 + 0.03 + …

7/6 = 1 + 0.1 + 0.06 + 0.005 +... = 1.165 (approx)

Personally I feel second method is better as it’s a generic way of tackling fractions and percentages. You just need to brush up your addition skills on two and three digit numbers. Represent numerator in terms of denominator. That’s all it takes.

321/562 = ____

321 = 281 (562/2) + 28.1(562/20) + 11.24(562/50) + …

321/562 = 0.5 + 0.05 + 0.02 + … = 0.57(approx)

562/321 = ___

562 = 321 + 160.5 (321/2) + 80.25 (321/4) + …

562/321 = 1 + 0.5 + 0.25 = 1.75 (approx)

You will find these methods very useful while dealing with calculation intensive DI problems or number system questions in quant.If you feel this method is helping (or can help) to save some time in calculation, practice well so that these stuffs will come automatically when you solve a question. Reduce your pen usage just to write the sub values of an expression and not to write the steps.

Here we are not using any speed math techniques, but we are trying to do smart math. Forget the “magic trick” of finding the square of 4 digit numbers by drawing circles or finding the square root by connecting dots. Stick to the basics and make it strong which will be good enough for kind of calculations expected in our aptitude exams. Everything we need to win an aptitude exam is taught to us when we were in high school... We just need to brush it up.

There will be lot more techniques which comes under the criteria **Simple to learn & easy to use**. Please do share your secret tricks... You can publish your gyan notes or can post your tricks as comments here.

**Happy Learning!** **:)**

#### Anatomy of Numbers

Winning an aptitude test takes more than just marking the correct options. Given enough time, required cutoffs are very much doable. What matters most is the approach we choose to reach our solutions. Even a question marked correctly can be a wrong choice if it wasted time that could have been more productive if spent on other questions. Also an unattended question can bring a smile to a smart test taker if the question was unreasonably demanding.

As a rule of thumb, always spend about 10 seconds to contemplate the best approach before start solving a question. Quant in general and number systems in particular is fun once we understand the ‘anatomy’ of numbers. A clear understanding on this area shall help in breaking down complex calculations and also help us to simplify the expressions which can then be solved easier than the conventional methods.

**Number line**

A number line is a straight line on which every point is assumed to correspond to a real number and every real number to a point.

This is a simple example of coordinate system. An arbitrary point O (the origin) is chosen on a given line. The coordinate of a point P is defined as the signed distance from O to P, where the signed distance is the distance taken as positive or negative depending on which side of the line point P lies. All real numbers can be represented in this line.

From a number line perspective if X needs to reach point 29.5 from origin, X has to ‘JUMP’ 2 steps each of 10 units then another 9 steps of 1 unit each then another 5 steps of 1/10 unit each. Or X can just make 3 steps of 10 units and then ‘COME BACK’ 5 steps each of 1/10 units. Or he can JUMP one step of 1 unit and then ‘REPEAT’ the same for 29 times then JUMP a step of 1/10 unit and REPEAT the same for 5 times… in all the above cases X is going to be @ 29.5

When we write the above mumbo jumbo in math it looks familiar

0 + 10 + 10 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 = 29.5

0 + 10 + 10 + 10 – 0.1 -0.1 -0.1 -0.1 -0.1 = 29.5

0 + 1 * 29 + 0.1 * 5 = 29.5

In our number system questions if you see Mr. X doing lot of JUMPING or SKIPING or FLYING, don’t panic. After this circus, X will settle in a POINT and we will figure it out.

**Place Value**

When we write numbers, each digit has a place value.

Consider number 123

3 is Units position means place value is 1. We have 3 Ones

2 is Tens position means place value is 10. We have 2 Tens

1 is Hundreds position means place value is 100. We have 1 Hundreds

123 = 1 * 10^{2} + 2 * 10^{1} + 3 * 10^{0}

**Decimal point**

Concept of decimal point helps us not to limit math just to integers, but to extend to infinitely more possibilities between the integers in number line. As we move further left of decimal point, every number place gets 10 times bigger. As we move further right, every number place gets 10 times smaller (one tenth as big).

123.123 = 1 * 10^{2}+ 10^{1}* 2 + 10^{0}* 3 + 1 * 10^{-1}+ 10^{-2}* 2 + 10^{-3}* 3

**Base System**

When we wrote123 = 1 * 10^{2}+ 10^{1}* 2 + 10^{0}* 3, we expressed 123 in terms of powers of 10. As we go left the place value increased 10 times and as we go right place value reduced 10 times. We are representing 123 ‘BASE’d on 10. We can say number 123 is represented in base 10, also called as decimal system.

**Algebraic representation of numbers**

Representing numbers in algebraic form is very useful while we chase X. Sharing some useful Fundas

Consecutive numbers: n, n+1, n+2 …

Even number: 2n

Consecutive even numbers: 2n, 2n+2, 2n+4 …

Odd number: 2n+1

Consecutive Odd numbers: 2n+1, 2n+3, 2n+5 …

2 digit number (say ab) = 10a + b (a and b can take values from 0 to 9)

3 digit number (say abc) = 100a + 10b + c

n digit number =10^{(n-1)}digit_{1}+ 10^{(n-2)}digit_{2}+… + digit_{n }(Digits taken left to right)

Three consecutive numbers: n-1, n, n+1 (sums to zero)

(a + b)^{2}= a^{2}+ 2ab + b^{2}

(a - b)^{2}= a^{2}- 2ab + b^{2}

(a + b)(a - b) = a^{2}- b^{2}

a^{3}+ b^{3}= (a + b) (a^{2}- ab + b^{2})

a^{3}- b^{3}= (a - b) (a^{2}+ ab + b^{2})

(a + b)^{3}= a^{3}+ 3a^{2}b+ 3ab^{2}+ b^{3}

(a - b)^{3}= a^{3}- 3a^{2}b+ 3ab^{2}- b^{3}

(a + b + c)^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca

(a + b + c) (a^{2}+ b^{2}+ c^{2}- ab - bc - ca) = a^{3}+ b^{3}+ c^{3}- 3abc

a^{3}+ b^{3}+ c^{3}= 3abc, if a + b + c = 0

a^{0}= 1

a^{1}= a

a^{m}× a^{n}= a ^{(m + n)}

a^{m}/ a^{n}= a ^{( m – n )}

(a^{m})^{n}= a^{m.n}

a^{- m}= 1 / a^{m }

(ab)^{m}= a^{m}b^{m}

Symbol √ is called as radical sign or radix.

Need a case to solve? Here we go…

**When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed? (CAT 2006)**

Here we are not just asked to find X but the whole X Gang! What are the clues?

1. X Gang contains only two digit numbers

2. If X Gang members are reversed they are increased by 18 than the original value.

We can write any 2 digit number xy as 10x + y (x and y are the first and second digit).

After reversing the number is yx represented as 10y + x.

Given 10y + x = 10 x + y + 18, solving y = x + 2. y should be 2 more than x.

Now y cannot be 2 as then x = 0 and the number 02 is not a 2 digit number!

y can take values from 3 to 9

when y = 3, x = 3 – 2 =1; y=4, x = 4 – 2 = 2; and so on…

Our gang is (10x + y) for all the above (x, y) which are 13, 24, 35, 46, 57, 68 and 79.

Gang busted! :)

**Simplify**

Before writing complicated expressions into work sheet, always remember, X can take 25 steps forward and 24 steps backward, which is as good as taking one step forward. If we take the route X ‘wants us to take’ we will take 49 steps than the required 1 step! While attending quant never jump into solving. Read the question carefully, understand the clues, ideate where X can be, form a strategy that budgets our time and efforts in best possible way and then start solving.

Question makers CAN and WILL include traps that can mislead us to unnecessary routes wasting precious time. Beware!

**Divide and Conquer**

We will now learn an interesting and equally important concept, divisibility of numbers. These concepts will be used extensively during prime factorization and while calculating HCF/LCM. Here we will just focus on how we can easily check whether a given number is divisible by some common divisors or not.

**Divisible by 2**: If the last digit is divisible by 2.

12, 142, 68…

**Divisible by 3**: Sum of digits of the number is divisible by 3.

15672, sum of digits = 1+5+6+7+2 = 21 = 3 * 7, hence divisible by 3.

**Divisible by 4**: If the last 2 digits are divisible by 4.

724, Last 2 digits (24) gives a number divisible by 4.

**Divisible by 5**: If the last digit is 5 or 0.

E.g. 625, 310 etc…

**Divisible by 7**: Subtract twice the unit digit from the remaining number.

If the result is divisible by 7, the original number is.

14238, 2 * 8 – 1423 = -1407 = -201 * 7, hence divisible by 7

**Divisible by 8**: If the last 3 digits are divisible by 8.

1040, Last 3 digits (040) gives a number divisible by 8.

A number is divisible by 2^{n}if the last n digits are divisible by 2^{n}.

**Divisible by 9**: If the sum of the digits is divisible by 9

972036, sum of the digits = 9 + 7 + 2 + 0 + 3 + 6 = 27, divisible by 9.

**Divisible by 11**: If the difference between the sum of digits at the odd place and the sum of digits at the even place is zero or divisible by 11.

1639, (9+6) - (3+1) = 11, divisible by 11.

**Divisible by 13**: If the difference of the number of its thousands and the remainder of its division by 1000 is divisible by 13.

2184, 2 - 184 = -182, so divisible by 13.

**Divisible by 33, 333, 3333… & 99, 999, 9999…:**

As a general rule, to check a given number is divisible by 333…3 (n digits) just see whether the sum of digits taken n at a time from right is divisible by 333…3 (n digits). If yes then the original number is also divisible by 33…3 (n digits). It is easy to understand through examples.

Is 627 divisible by 33?

Take 2 digits from right at a time, and get the sum.

27 + 06 = 33, hence divisible by 33

Is 22977 divisible by 333?

Take 3 digits from right at a time and find the sum.

977 + 022 = 999, hence divisible by 333

Same can be applied for checking the divisibility of a given number with 99…9 (n digits). Check if the sum of digits taken n at a time from right is divisible by 999…9 (n digits). If yes then the original number is also divisible by 99…9 (n digits)

Is 6435 divisible by 99?

35 + 64 = 99. As per the above rule, 6435 is divisible by 99.

How much time you need to tell whether the number 1000000998 is divisible by 999?

All Math enthusiasts out there, can you try to figure out the reason behind this curious behavior of 33…3 (n digits) and 99…9 (n digits) using number line and base concept? :)

**Some practice exercise**

1. 32X9 is divisible by 11 then A = ___

2. Is the number 156573 divisible by 24?

3. 8X36Y is divisible by 4, 8 and 11 find the number.

4. What is the reminder when 56784 is divided by 91?

5. Is the number 11632 divisible by 16?

**Solutions: **

1. 9 + 2 – X – 3 = n * 11, possible only when X = 8. Number is 3289

2. No. Check for divisibility with 8 and 3.

3. Y can take 4, 8 to be divisible by 4. But number is divisible by 8 only when Y = 8

8 + 3 + 8 – 6 - X = n * 11 (Divisibility by 11) possible only when X = 2. Number is 82368.

4. Zero. 91 = 13 * 7, the given number is divisible by 13 and 7.

5. Yes. Last 4 digits are divisible by 16.

**Approximate**

Wherever possible, approximate. Most of the times we don’t need precision level more than 2 decimal points to uniquely differentiate the given options. If the question demands more precision than that, leave the question and come back if you have time.

As we discussed earlier, an easy approximation technique is to write numerator with respect to denominator

2136 / 17 =

2136 ≈ 1700 (100 * 17) + 340 (20 * 17) + 85 (17 * 05) + 8.5 (17 * 0.5) + 1.7 (17 * 0.1) …

2136 / 17 ≈ 100 + 20 + 5 + 0.5 + 0.1 +… ≈ 125.6 ( Actual value is 125.64 )

There are various approximation techniques which are discussed and debated at multiple forums. But I feel most of them are complicated considering the level of approximation we need. Stick with methods that are simple to understand and easy to apply. I am not sure, but many approximation techniques are some where related to the aforementioned method. If you know some methods which are useful in approximation, do share.

**Use Options **

**Number S is equal to the square of the sum of the digits of a 2 digit number D. If the difference between Sand D is 27, then D is (CAT 2002)**

(1) 32

(2) 54

(3) 64

(4) 52

Here we can form the algebraic equation and solve… but easiest way is to take options one by one and see which option satisfies the given condition.

Option1: 32, S – D = (3 + 2)^{2}– 32 = 7, Not the answer.

Option2: 54, S – D = (5 + 4)^{2}– 54 = 27… Oila!!! :)

Options are not just useful in substitution, but will also help us in deduction also.

‘Solving’ this equation will take us ages. We can write the above question as

Options can be written as

(1) (n+1) – 1/ (n+1)

(2) n – 1/n

(3) n – 1/ (n+1)

(4) (n+1) – 1/n

(5) (n+1) – 1/ (n+2)

For n =1, sum is √ (2+1/4) = √9/4 = 3/2

Now substitute n=1 in the options

1) 2 – 1/2 = 3/2

2) 1 – 1/1 = 0

3) 1- 1/2 = 1/2

4) 2- 1/1 = 1

5) 2 – 1/3 = 5/2

Only option 1 satisfies. This holds true for any value of n. Coming back to the original question where n = 2007, sum should be equal to (n+1) – 1/ (n+1) = 2008 – 1/2008.

**Happy learning! :)**

#### Factors of a Number

A divisor of an integer n, also called a factor of n, is an integer which divides n without leaving a remainder. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words a composite number is any positive integer greater than one that is not a prime number.

Literally prime means ‘of first importance’ and composite means ‘made up of various parts or elements’. Now what a composite number made up of? Prime numbers! **All composite numbers can be written as a product of prime factors. This is called as the standard form of a composite number. **We can say thatprime numbers are the basic building blocks of all numbers.

For example, 42 can be written as 2 * 3 * 7 and 50 can be written as 2 * 5^{2}

Always keep in mind that 1 is neither prime nor composite. But why 1 is neither prime nor composite?

In number theory, the fundamental theorem of arithmetic states that every integer greater than 1 is either prime itself or can be **uniquely** written as a product of prime numbers. Take the case of 90. Theorem states that 90 is either a prime or a product of primes. By factorization, we get 90 as product of primes. 90 = 2 * 3* 3 * 5. It also says that the combination of primes will be unique (if we disregard the order of primes). Means 90 can be written as 5 * 3 * 3 * 2 or 3 * 2 * 3 * 5 or any combination we can form using the prime. But there will be always one 2, two 3s and one 5 in the expression, which is unique to 90. If we treat 1 as a prime, then we can bring in any number of ones. We can write 90 as 5 * 3 * 3 * 2 * 1 and 5 * 3 * 3 * 2 * 1 * 1 which are both valid expression for 90 and this violate the unique factorization requirement of fundamental theory. So one is not considered as a prime and as one cannot be represented as a product of primes and hence one is not composite too.

Zero is neither a prime number nor a composite number. It cannot be a prime because it has an infinite number of factors and cannot be composite because it cannot be expressed by multiplying prime numbers.

We can get lot of information related to a number from its standard form. Here we go

**Number of factors of a number**

**If a number N can be represented as P _{1}**

^{a}*** P**

_{2}

^{b}*** P**

_{3}

^{c}**where P**

_{1}, P_{2}, P_{3}are prime factors of N then number of factors of N can be found as (a + 1) * (b + 1) * (c + 1).50 = 2 * 5^{2}, so number of factors of 50 = (1+1) * (2+1) =6. Which are 1, 2, 5, 10, 25and 50. These are various combinations of the prime factors 2, 5 and 5.

Similarly, 792 = 2^{3}* 3^{2}* 11, number of factors = (3+1) * (2+1) * (1+1) = 24

You can see how extensively we use divisibility rules here. For the above problem it won’t take much time to see the number 792 has a factor 2 (ending with even number), factor 4 (last 2 digits divisible by 4), factor 8 (last 3 digits divisible by 8 ), factor 3 (sum of digits divisible by 3), factor 9 (sum of digits divisible by 9) and factor 11 (difference between sum of digits at odd place and sum of digits at even place is zero or multiple of 11, here 0).

To find the standard form divide the number with its prime factors (take highest power of each prime) till we get unity. Then group the powers of all prime factors we used in each step to get the standard form. Confused? It is easy with examples J

Take 544; By using the divisibility rules we can easily see 2, 4, 8 are factors of 544. Take the highest power of each prime (here 8 ) for division.

544/8 = 68, divisible by 2, 4 and 17. Take the highest power of each prime

68/ (4 * 17) = 1.

544 = 8 * 4 * 17 = 2^{5 }* 17

Number of factors = 6 * 2 = 12

Need one more? Take 1080; Applying divisibility rules, we get 2, 3, 4, 5, 8, 9 divides the number. Take the highest power of each prime (here 8, 5, 9) for division.

1080 / (8 * 5 * 9) = 3 (can stop dividing here as we got a prime, no need to write an obvious step to get unity by dividing with the same prime!)

1080 = 8 * 5 * 9 * 3 = 2^{3}* 3^{3}* 5

Number of factors = (3 + 1) * (3 + 1) * (1 + 1) = 4 * 4 * 2 = 32

**Sum of factors of a number**

40 = 2^{3} * 5, number of factors = (3 + 1) * (1 + 1) = 4 * 2 = 8.

Sum of the factors = 1 + 2 + 4 + 8 + 5 + 10 + 20 + 40 = 90

This is same as (2^{0} + 2^{1} + 2^{2} + 2^{3}) (5^{0} + 5^{1}) = 15 * 6 = 90.

Find the sum of all the powers of each prime numbers, starting from 0 to the highest power contained in the standard form. Product of all such sums will give us the sum of the factors.

Here also it is easy to understand with examples, so here we go.

We have already seen that 1080 = 2^{3} * 3^{3 }* 5

In this case, sum of the factors will be (2^{0 }+ 2^{1} + 2^{2} + 2^{3}) * (3^{0 }+ 3^{1} + 3^{2} + 3^{3}) * (5^{0} + 5^{1})

= 15 * 40 * 6 = 3600

This also explains how we find the numbers of factors. It is the product of the number of factors in each bracket (here, 4 * 4 * 2 = **32**).

**Product of factors of a number**

N = P_{1}^{a} * P_{2}^{b} * P_{3}^{c }

Then the product of all the factors of N = N ^{(a + 1) (b + 1) (c + 1)/2}

e.g. 50 = 2 * 5^{2}; Product of all factors of 50 = 50 ^{(6) / 2} = 50^{3}

**Number of odd factors**

To get the number of odd factors, ignore the powers of 2.

Take 1080, 1080 = 2^{3} * 3^{3} * 5

Number of odd factors of 1080 = (3 + 1) * (1 + 1) = 4 * 2 = 8, here we considered only the powers of 3 and 5 to get the number of odd factors.

**Sum of odd factors of a number**

Sum of factors of 1080 = (2^{0 }+ 2^{1}+ 2^{2}+ 2^{3}) * (3^{0}+ 3^{1}+ 3^{2}+ 3^{3}) * (5^{0}+ 5^{1}), where each term in the expansion represents a factor of 1080.

We need at least one 2 to get an even factor. Remove all terms that has a 2 in it and we will remove all even factors. Even factors in 1080 are 2^{1}, 2^{2}and 2^{3}. The only power of 2 remaining is 2^{0 }(=1), which can be removed as it will not make any impact. Means, you ignore all powers of 2 to get the sum of all odd factors.

Now we need to remove even factors from the expression so that the result will be the sum of all odd factors.

Sum of odd factors of 1080 = (3^{0}+ 3^{1}+ 3^{2}+ 3^{3}) * (5^{0}+ 5^{1}) = 240

**Number of even factors of a number**

Let a number is of the form, N = 2^{a} * P_{2}^{b} * P_{3}^{c}…

Then Number of even factors of N = a * (b + 1) * (c + 1) …

For e.g. 1080 = 2^{3 } * 3^{3} * 5

Number of even factors of 1080 = 3 * (3 + 1) * (1 + 1) = 24

**Sum of even factors of a number**

To get the sum of the even factors we need to ensure that all the terms in the expression for the sum of factors are even. All members in the 2’s bracket except 2^{0}will give an even number. Remove 2^{0 }from the group and that’s all it takes!

For e.g. 1080 = 2^{3 } * 3^{3} * 5

Sum of even factors of 1080 = (2^{1} + 2^{2 }+ 2^{3}) * (3^{0 }+ 3^{1} + 3^{2 }+ 3^{3}) * (5^{0 }+ 5^{1}) = 3360

**Other conditions**

What if we are asked to find the sum and number of factors of 1080 which are divisible by 15?

We don’t have to use anything else other than the logic we used for odd and even factors. We know number of terms in the expression is the number of factors. So when a condition is given, number of factors is the number of terms in the expression satisfying the given condition and considers only those factors to get the sum.

1080 = (2^{0 }+ 2^{1}+ 2^{2}+ 2^{3}) * (3^{0}+ 3^{1}+ 3^{2}+ 3^{3}) * (5^{0}+ 5^{1})

Now 15 = 3 * 5, so we need to ensure that every term in the expression has a 3 and 5 in it. Now look at the bracket of 3 and find the terms that don’t yield a 3 in the expression. Also check for bracket of 5 for the terms that don’t give us a 5. 3^{0}and 5^{0}are the culprits. Remove those entries and we get the expression we need.

**Sum of factors of 1080 which are divisible by 15**

= (2^{0 }+ 2^{1}+ 2^{2}+ 2^{3}) * (3^{1}+ 3^{2}+ 3^{3}) * (5^{1})

= 15 * 39 * 5 = 2730

**Find the sum of factors of 544 which are perfect squares**

544 = 8 * 4 * 17 = 2^{5 }* 17

Our expression, (2^{0}+ 2^{1}+ 2^{2}+ 2^{3}+2^{4 }+2^{5}) (17^{0}+17^{1})

Ensure all terms in the expression satisfy the condition, perfect squares. So each term in the expression should be of even power. We have to remove all odd powers in the expression

Required sum is (2^{0}+ 2^{2 }+2^{4}) (17^{0}) = 21.

Number of factors = 3 * 1 (number of terms in each bracket) = 3.

**For the number 1000 find the below values**

**The number of divisors **= (3 + 1) * (3 + 1) = 16 (as 1000 = 2^{3}* 5^{3})**The product of divisors** = 1000^{16/2 }= 1000^{8}**The sum of divisors**= (2^{0}+ 2^{1}+ 2^{2}+ 2^{3}) (5^{0 }+ 5^{1}+ 5^{2}+ 5^{3}) = 2340

**The sum and number of odd divisors**

we need to remove all factors that **can** yield a 2 from the two’s bracket.

Sum of odd factors = ( 2^{0}) * (5^{0}+ 5^{1}+ 5^{2}+ 5^{3}) = 156

Number of odd factors = 1 * 4 = 4

**The sum and number of even divisors**

We need to remove all factors that **cannot** yield a 2 from the two’s bracket (which is 2^{0})

Sum of even factors = (2^{1}+ 2^{2}+ 2^{3}) (5^{0}+ 5^{1}+ 5^{2}+ 5^{3}) = 2184

Number of even factors = 3 * 4 = 12

**The sum and number of divisors which are perfect squares**

We need to remove all the factors which are not perfect squares

Sum of factors which are perfect squares = (2^{0}+ 2^{2 }) (5^{0}+ 5^{2}) = 130

Number of factors which are perfect squares = 2 * 2 = 4

**The sum and number of divisors which are not perfect squares**

Total sum – sum of divisors which are perfect squares = 2340 – 130 = 2210

Total factors – number of divisors which are perfect squares = 16 – 4 = 12

**The sum and number of divisors which are divisible by 125**

We need to remove all the factors that cannot yield 125 (125 = 5^{3})

Sum of factors which are divisible by 125 = (2^{0}+ 2^{1}+ 2^{2}+ 2^{3}) (5^{3}) = 1875

Number of factors which are divisible by 125 = 4 * 1 = 4

**The sum and number of divisors which are divisible by 100**

100 = 2^{2 }* 5^{2 }

We need to remove all factors that cannot yield a 2^{2 }from two’s bracket and also remove all factors that cannot yield a 5^{2 }from five’s bracket.

Sum of factors which are divisible by 100 = (2^{2}+ 2^{3}) (5^{2}+ 5^{3}) = 1800

Number of factors which are divisible by 125 = 2 * 2 = 4

**Co Primes**

We will end this chapter with a very useful concept, co primes. A co prime is a number which has no common prime numbers in their standard form. 20 = 2^{2 }* 5 and 21 = 7 * 3, no common primes hence 20 and 21 are co primes. (Two consecutive integers will always be co prime)

The number of integers co prime to a positive integer n, within the range [1, n-1], is given by Euler's phi function (phi)

If a number N = P_{1}^{a} * P_{2}^{b }* P_{3}^{c} then phi(N) = N ( 1 - 1/ P_{1}) ( 1 - 1/ P_{2}) ( 1 - 1/ P_{3})

Consider 6, 6 = 2 * 3

Phi (6) = 6 (1 – 1/2) (1 – 1/3) = 6 *1/2 * 2/3 = 2, means there are 2 numbers between 1 and 6 which are co prime to 6. (1 and 5). The numbers 1 and -1 are coprime to every integer, and they are the only integers to be co prime with 0.

Take another example, 24; 24 = 2^{3} * 3

phi(24) = 24 ( 1 – 1/2) (1 – 1/3) = 24 * 1/2 * 2/3 = 8. There are 8 numbers between 1 and 24 which are co prime to 24. (1, 5, 7, 11, 13, 17, 19, 23)

One more example, 36; 36 = 2^{2 }* 3^{2}

phi(36) = 36 ( 1 – 1/2) (1 – 1/3) = 36 * 1/2 * 2/3 = 12 . There are 12 numbers less than 36 and co prime to 36. (1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35)

If a and b are relatively prime, then

- 2
^{a}-1 and 2^{b}-1 are also relatively prime. - There exist integers x and y such that ax + by = 1 (Bézout’s identity)
- Any powers a
^{j}and b^{k}are also co prime.

**Happy Learning :-)**

#### HCF & LCM fundas

HCF and LCM are among the initial ‘complicated’ stuffs we dealt with as a student. Someone taught me once that HCF is a postman who can go and meet all of the given numbers to deliver their letters. And LCM is like a postmaster to whom all the given numbers can go in case of any issue. Highest common factor (HCF), also known as GCD (Greatest common divisor) of a group of numbers is the greatest number that can exactly divide all the numbers in the group. Least Common Multiple (LCM) of a group of numbers is the smallest number which can be divided by all the numbers in the group.

**Finding HCF of given numbers**

**Factorization Method**

**Find the HCF of 100, 225 and 900**

Represent each number in their standard form.

100 = 2^{2 }* 5^{2}

225 = 3^{2} * 5^{2}

900 = 2^{2} * 3^{2} * 5^{2}

what is common in all of them, 5^{2}. HCF (100, 225, 900) = 25.

So 25 is the largest integer that can completely divide 100, 225 and 900.

**Division Method:**

To find the HCF of two given numbers, divide the largest by the smallest number, and then divide the dividend by the remainder. Repeat this until remainder is zero. The last dividend is the HCF of the two numbers.

**Find HCF of 27 & 36**

Step1: Divide 36/27, remainder = 9

Step2: Divide dividend by remainder

27/9, remainder = 0

So the HCF or GCD (greatest common Divisor) of 27 and 36 is 9**To find the HCF of three given numbers**

**Calculate the HCF of 2 numbers, then HCF of the result with 3rd number.**

**Find the GCD of 100, 225, & 900 by Successive Division Method**

**Phase1**: Find GCD of two numbers (here 900 and 100)

Step 1: Divide the largest number 900 by the smallest number 100. And this division gives us a remainder 0 and the divisor is the GCD of 100 and 900 = 100.**Phase2**: Find the GCD of 100(GCD of 900 & 100) & 225(given number)

Step 1: Divide the remaining given number 225 by the GCD of 100 & 900 i.e.100.

Step 2: Division in Step 1 gives us remainder 25. Divide the dividend (100) again by remainder (25), which gives us a remainder 0 and the divisor (25) is our GCD.

GCD of 100, 225 and 900 = 25

**Finding LCM of given numbers**

**Factorization method**

**Find the LCM of 100, 225 and 900**

100 = 2^{2 }* 5^{2}

225 = 3^{2 }*5^{2}

900 = 2^{2 }* 3^{2 }* 5^{2}

Write down all the prime factors that appear at least once in any of the number and then raise it to their highest available power. Product of them will give you the LCM. Using this logic in our above problem, LCM = 2^{2 }* 3^{2 }* 5^{2} = 900. Which means 900 is the smallest number which can be completely divided by the numbers 100, 225 and 900.

**Division method**

Write the given numbers as shown below and divide them with the prime numbers that can exactly divide one or more of the given numbers. On division, write the quotient in each case below the number. If any number is not divisible by its respective divisor, it is to be written as such in the next line. Keep on dividing the quotient until you get 1(as quotient of all). Multiply all the divisors to get LCM of given numbers.

**Find LCM of 100, 225 and 900**

5 | 100 225 900

5 | 20 45 180

2 | 4 9 36

3 | 2 3 6

2 | 1 3 3

3 | 1 1 1

LCM (100, 225, 900) = 3 * 2 * 3 * 2 * 5 * 5 = 900.

We can extend division method to find GCD or LCM of any given numbers. It is much easier and time saving to use the prime factorization method as once we write the numbers in their standard form we can get the answer just by visual inspection. Only thing we need is the speed and accuracy to write the standard form of the number, which comes through practice. In case of division method for LCM a closer look will show you that we are just using standard form method in another way.

**LCM for comparing fractions**

Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

Which one is greater 10/9 or 7/6?

LCM of denominator = 18.

Make denominator same as the LCM for the both the fractions

20/18, 21/18 second one has the highest numerator hence the highest fraction is 7/6.

**HCF and LCM of decimal numbers**

In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

Find the HCF and LCM of 0.12 and 0.3

Add one zero to 0.3 so that we have same number of decimal places, 0.12 and 0.30, consider these numbers without decimals i.e. 12 and 30 and find HCF (6) and LCM (60). Mark as many decimals we removed (here 2). So the answer is HCF = 0.06 and LCM = 0.6

**Useful concepts **

HCF (n_{1,} n_{2}) * LCM (n_{1,} n_{2}) = n_{1 }* n_{2 }(works only for TWO numbers)

LCM of fractions = LCM of numerators/HCF of denominators.

HCF of fractions = HCF of numerators/LCM of denominators.

Two numbers are co primes if their HCF is 1.

LCM is always greater than or equal to the biggest among the given numbers.

HCF is always less than or equal to the smallest among the given numbers.**Some regular scenarios asked from HCF / LCM **

Find the GREATEST NUMBER that will exactly divide x, y, z.

Required number = H.C.F. of x, y and z.

Find the GREATEST NUMBER that will divide x, y & z leaving remainders a, b and c respectively.

Required number = H.C.F. of (x – a), (y – b) and (z – c).

Find the LEAST NUMBER which is exactly divisible by x, y and z.

Required number = L.C.M. of x, y and z

Find the LEAST NUMBER which when divided by x, y and z leaves the remainders a, b and c respectively. It is always observed that (x – a) = (z – b) = (z – c) = K (say).

Required number = (L.C.M. of x, y and z) – K.

Find the LEAST NUMBER which when divided by x, y and z leaves the same remainder ‘r’ each case.

Required number = (L.C.M. of x, y and z) + r.

Find the GREATEST NUMBER that will divide x, y and z leaving the same remainder in each case.

Required number = H.C.F of (x – y), (y – z) and (z – x).

Try out the above formulae by taking some arbitrary numbers and also try to understand why it works. And most importantly share them too.In our exams, we may not find a problem directly asking to find the HCF and LCM. But questions where we need to apply the concept of HCF and LCM are very common.

**The front wheels of a toy truck are 4 inches in circumference. The back wheels are 7 inches in circumference. If the truck travels in a straight line without slippage, how many inches will the truck have traveled when the front wheels have made 12 more rotations than the back wheels?**

(A) 112 (B) 64 (C) 48 (D) 36 (E) 28

Front wheel covers 4 inches in one rotation and back wheel covers 7 inches for one rotation. Distance travelled by the truck will be an integral multiple of the LCM of 4 and 7 (distance covered by front wheel and back wheel for each of their rotation). LCM of 4 and 7 is 28 (4 and 7 are co primes hence HCF is 1. We can get LCM by multiplying the given numbers as HCF * LCM = Number1 * Number2) Only options that can be written as 28n are 112 and 28.

If 28, front wheel will make 7 rotations and back wheel will make 4 rotations.

Difference is 3. Not our answer.

If 112, front wheel will make 28 rotations and back wheel will make 16 rotations.

Difference is 12… our answer :)

If options are not available, then we can see that for each 28 inch (LCM) covered by the truck the difference between the number of rotations made by the front wheel and back wheel will be 3 more than the previous value, which is nothing but the difference of the given numbers divided by their HCF (here HCF = 1). Means for the first revolution, front wheel will make 7 rotations and backwheel will make 4 rotations. Difference is 3… for the second revolution difference will be 6 and so on… going with this trend after the truck covers 28 inches for fourth time the difference will be 4 * 3 = 12. So the required distance is 4 * 28 = 112.

**Happy Learning :-)**

#### Factorials

Factorial is an important topic in quantitative aptitude preparation not just because of its importance as an independent concept, but also because of its application in many topics like permutation & Combination, Probability etc… The factorial of a non negative integer n is denoted as n! The notation was introduced by Christian Kramp in 1808.

n! is calculated as the product of all positive integers less than or equal to n.

i.e 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720

n! = 1 when n = 0, and n! = (n-1)! * n if n > 0

**What is the significance of n!**

**n! is the number of ways we can arrange n distinct objects into a sequence.**

2! = 2 means numbers 1, 2 can be arranged in 2 sequences (1, 2) and (2, 1).

We can arrange 0 in one way. So 0! = 1, not zero. Now we know why, and no need to say “its like that” if someone asks ;-)

**Find the highest power of a prime number in a given factorial**

The highest power of prime number p in n! = [n/p^{1}] + [n/p^{2}] + [n/p^{3}] + [n/p^{4}] + …

[x] denotes the greatest integer less than or equal to x.

[1.2] =1

[4] = 4

[-3.4] = -4 (why? Because -4 is less than -3.2)

**Find the highest power of 3 in 100!**

= [100/3] + [100/3^{2}] + [100/3^{3}] + [100/3^{4}] + [100/3^{5}] + …

= 33 + 11 + 3 + 1 + 0 (from here on greatest integer function evaluates to zero)

= 48.

**What is the greatest power of 5 which can divide 80! exactly** (Another way of asking what is the highest power of 5 in 80!)

= [80/5] + [80/5^{2}] + [80/5^{3}] + …

= 16 + 3 + 0 + … = 19.

**Find the highest power of a non prime number in a given factorial**

We learnt before that any non prime number can be expressed as a product of its prime factors. We will use this concept to solve this type of questions.

**What is the greatest power of 30 in 50!**

30 = 2 * 3 * 5.

Find what the highest power of each prime is in the given factorial

We have 25 + 12 + 6 + 3 + 1 = 47 2s in 50!

16 + 5 + 1 = 22 3s in 50!

10 + 2 = 12 5s in 50!

We need a combination of one 2, one 3 and one 5 to get 30. In the given factorial we have only 12 5s. Hence only twelve 30s are possible. So the greatest power of 30 in 50! is 12.

**Note: We don’t have to find the power of all prime factors of a given number. It is enough to find the greatest power of the highest prime factor in the factorial (here 5).**

**What is the greatest power of 8 in 50!**

8 =2^{3}

50! has 47 twos. We need 3 two to get an 8.

We can have a maximum of [47/3] = 15 8s in 50!

**Note: The highest power of the number p**^{a}** in n! = [Highest power of p in n!]/a**

**n! in its standard form**

**Find the highest power of all prime numbers, less than n, in n!**

**What are the number of factors of 15! **

2, 3, 5, 7, 11 and 13 are the prime numbers less than 15.

Highest power of 2 in 15! = 11

Highest power of 3 in 15! = 6

Highest power of 5 in 15! = 3

Highest power of 7 in 15! = 2

Highest power of 11 in 15! = 1

Highest power of 13 in 15! = 1

15! = 2^{11 }* 3^{6 }* 5^{3 }* 7^{2}* 11^{1 }* 13^{1}

Thus we can find the number of divisors, sum of divisors and other values for 15!

number of factors of 15! = (11 +1) (6 + 1) (3 + 1) ( 2 + 1) ( 1 + 1) (1 + 1) = 4032

**Find the number of zeros at the end of a factorial value**

This is just another way of asking what highest power of 10 is in a given factorial.

10 = 2*5, we need a combination of one 2 and one 5 to get a zero at the end. Just find the number of 5s in the given factorial as number of 2s will be always more than number of 5s.

**Find the number of zeros at the end of 100!**

[100/5] + [100/25] + … = 20 + 4 + 0 + … = 24.

**Find the number of zeros at athe end of 100! in base 7**

Concept is same as before. in base 10 we need a 2 and 5 to get a zero but in base 7 we need a 7 to get a zero. so we have to find the highest power of 7 in 100!

[100/7] + [100/49] + ... = 14 + 2 = 16. There will be 16 zeros at the end of 100! in base 7

**Note: To find the number of zeros of a factorial in a base b, calculate the highest power of b in that factorial. b need not be prime, we already know how to find the highest power of a composite number in a factorial expression.**

So how many trailing zeros are there in 500! in base 9 ?

**Wilson's theorem**

**For every prime number p, (p-1)! + 1 is divisible by p**

take p =7, then as per Wilson's theorem 7 is prime only if (6!) + 1 is divisible by 7 and it is :-)

some useful results derived from Wilson's theorem

**Remainder [ (p-1)!/p] = p - 1 ( where p is a prime number )**

Remainder [ 6!/7 ] = 7; Remainder [100!/101] = 100 and so on...

**Remainder [ (p-2)!/p] = 1 ( where p is a prime number )**

Remainder [ 5!/7 ] = 1; Remainder [99!/101] = 1 and so on...

**Some extra concepts**

**Product of n consecutive natural numbers is divisible by n! **

11 * 12 * 13 * 14 is completely divisible by 4!

This can be extended to obtain other interesting results.

consider 1 * 2 * 3 * 4 * 5 * 6 . as per above result 1 * 2 * 3 * 4 * 5 * 6 is divisible by 6!.

now take ( 1 * 2 * 3 ) * ( 4 * 5 * 6 ) . each of this is divisible by 3!

so 1 * 2 * 3 * 4 * 5 * 6 is completely divisible by (3!)^{2 }

similarly 1 * 2 * 3 * 4 * 5 * 6 is completely divisible by (2!)^{3}also.

**Note: To generalise (a*n)! is completely divisible by (n!)**^{a}**, where a and n are integers.**

**The product of all odd (even) integers up to an odd (even) positive integer n is called the double factorial of n. (though it has only about half the factors of the original factorial!). It is denoted as n!!**

9!! = 1 * 3 * 5 * 7 * 9 = 945

8!! = 2 * 4 * 6 * 8 = 384

**Product of first n! is called the super factorial of n, denoted as sf(n)**

sf(4) = 1! * 2! * 3! * 4!

**A factorion is a natural number that equals the sum of the factorials of its decimal digits.**

For example, 145 is a factorion because 1! + 4! + 5! = 1 + 24 + 120 = 145.

**Note: There are just four factorions (in base 10) and they are 1, 2, 145 and 40585.**

**A factorial prime is a prime number that is one less or one more than a factorial**

2 = 1! + 1, 3 = 2! + 1, 5 = 3! -1, 7 = 3! + 1 etc…

In the chapter Know your Numbers, we saw Triangular numbers where instead of multiplying the numbers, we added n consecutive numbers... Remember why it is called as Triangular numbers ?

The first triangular number is 1.

The second one is 1 + 2 = 3.

The third triangular number is 1 + 2 + 3 = 6 and so on.

We can calculate the nth triangular number using the formula **[(n) (n+1)]/2**

**Happy Learning :-)**

#### Get your BASE right

A main reason for our friction with quant is that we learn numbers... not number systems. Many years ago, when our life was not complex as today, basic representation of numbers was enough to handle our daily transactions. Don’t know when! But some point during our evolution someone felt the need to count and arrange things. May be some Cro-Magnon dude want to show off saying I caught 'more' fish or I am the 'tallest' in the group... and various number systems were developed to meet the purpose (No! not to abet the Cro-Magnon show off... for counting and arranging :) ) In this article we will see some interesting stuffs about numbers and number system.

Decimal system (base 10) is a positional number system which uses ten numeric digits (0, 1, 2 ... 9) to create any number we want. In a positional system, position of a digit is used to signify the power of its base that the digit is to be multiplied with. Decimal is a base 10 system means each position in a decimal number is BASEd on a power of 10. So we will start from zero and count till nine once we have one more than 9 rather than having a new numeric digit we say as have a 10 now (1 * 10^{1} + 0 * 10^{0}). The concept of using zero as a number and not merely a numeric symbol is attributed to India.

In a general way, a positional base-b numeral system (with b a natural number greater than 1 known as the radix), b basic symbols (or digits) corresponding to the first b natural numbers including zero are used. To generate the rest of the numerals, the position of the symbol in the figure is used. The symbols in the last position has its own value, and as it moves to the left its value is multiplied by b.To avoid confusion while using different number systems, the base of each individual number may be specified by writing it as a subscript of the number. If no base is mentioned treat it as a decimal system (base 10)

For example, when we write 1875 in decimal system we are saying there is one 10^{3}, eight 10^{2}, seven 10^{1} and five 10^{0}

During your exam if you see some questions like "In some island numbers are represented using base-8" don’t lose heart. it just means that in that island there are only 8 symbols ( 0 to 7) and after counting till 7 they will say 10 ( means 1 * 8^1 + 0 * 8^0.. which is equal to 8 in our decimal system). There is a general way to convert any base representation to a decimal number.

Number d_{1}d_{2}d_{3}....d_{n} in base b shall be converted to base 10 as d_{1} * b^{n-1 }+ d_{2} + b^{n-2 }+ ... + d_{n} * b^{0}

Convert 1432001 in base 5 to a decimal number

(1432001)_{5} = (1 * 5^{6} + 4 * 5^{5} + 3 * 5^{4} + 2 * 5^{3} + 0 * 5^{2} + 0 * 5^{1} + 1 *5^{0})

Now how will we convert a decimal number to another base representation? Before going to the details we know a base b system has b number of symbols to represent it numbers. Like base 2 has 2 (0 and 1), base 3 has three digits (0, 1, 2) and so on. Now a decimal system has 10 symbols (0 to 9) and another base say (base5) has only 5 symbols (0 to 4) so base 5 people don’t understand what is 5,6,7,8 or 9. So when we convert we cannot put any of these symbols to the base 5 number. Here is the trick. We will carry forward for every 5... That is 4 + 1 in base 5 is 10 (one 5 and zero 1). While discussing remainder concept we saw that when we divide a number with another number n, the possible remainder values are 0 to (n-1). Means division by 2 can yield only two remainders values, zero or one; division by 3 can give only zero, one or two as remainder and so on.

To convert a decimal number to another base, divide the given number with the base value... And then repeatedly divide the quotient with the base value till the quotient is zero. Write the remainder in each step and then use these remainders as place values and use the last remainder as most significant digit and the first one as the least significant digit.

To convert 461 to base 8

We get 461 = 715(base 8 )

Want to cross check..??? 715(base 8 ) = 7 * 8^2 + 1 * 8 + 5 * 8^0 = 461 :)

Convert 366 to binary (base 2)

Clear...??? Now convert 45.68 to base 2.

For the portion before decimal point we can do the same way we did before and we can get 45 = 101101. Now 0.68, instead of dividing, multiply with the base till you get the fractional part of the result as zero.

0.625 * 2 = 1.25 integer part is 1

0.25 * 2 = 0.50 integer part is 0

0.50 * 2 = 1.00 integer part is 1, fractional part is zero. We can stop :)

0.625 = 0.101

45.625 = 101101.101

How can we convert (101101.101)2 back to decimal...?

101101 the same way we did before...

1 * 2^{5 }+ 0 * 2 ^{4} + 1 * 2^{3} + 1 * 2^{2} + 0 * 2^{1} + 1 * 2^{0} =

32 + 0 + 8 + 4 + 1 = 45.

For decimal part, decrement the power by 1 as we go right from decimal point.

Here, 0.1010 = 1 * 2^{-1} + 0 * 2^{-2} + 1 * 2^{-3} + 0 * 2^{-4} = 0.5 + 0.125 = 0.625

101101.101 = 45.625

As an alternate method, in case of finite fractions, move the decimal point to the right end of the number. Count how many places we moved the decimal point. Convert the resulting integer to binary using normal method, and divide by 2^n, where n is the number of places we moved the decimal point.

For 101101.101,

101101101 (we moved decimal point to three places) = 365.

Now divide by 2^{3} = 8.

101101.101 = 365/8 = 45.625

We will see some other systems which were used to represent numbers. May be after reading about them you will fall in love with decimal number system :)

Finger numerals were used by the ancient Greeks, Romans, Europeans of the Middle Ages, and later the Asiatics.

The Mayan number system dates back to the fourth century and they used a base 20 system. Now the reason behind this is said as we have 20 fingers! The Mayan system used a combination of two symbols. A dot (.) was used to represent the units (one through four) and a dash (-) was used to represent five. For further study: The 360 day calendar also came from the Mayan's who actually used base 18 when dealing with the calendar. Each month contained 20 days with 18 months to a year. This left five days at the end of the year which was a month in itself that was filled with danger and bad luck. In this way, the Mayans had invented the 365 day calendar which revolved around the solar system.

The Egyptians used a written numeration that was changed into hieroglyphic writing, which enabled them to note whole numbers to 1,000,000. It had a decimal base and allowed for the additive principle. In this notation there was a special sign for every power of ten. This hieroglyphic numeration was a written version of a concrete counting system using material objects. To represent a number, the sign for each decimal order was repeated as many times as necessary. To make it easier to read the repeated signs they were placed in groups of two, three, or four and arranged vertically.

The Greek numbering system was uniquely based upon their alphabet. The Greek alphabet came from the Phoenicians around 900 B.C. When the Phoenicians invented the alphabet, it contained about 600 symbols. Those symbols took up too much room, so they eventually narrowed it down to 22 symbols. The Greeks borrowed some of the symbols and made up some of their own. But the Greeks were the first people to have separate symbols, or letters, to represent vowel sounds. Our own word "alphabet" comes from the first two letters, or numbers of the Greek alphabet -- "alpha" and "beta." Using the letters of their alphabet enabled them to use these symbols in a more condensed version of their old system, called Attic. The Attic system was similar to other forms of numbering systems of that era. It was based on symbols lined up in rows and took up a lot of space to write. This might not be too bad, except that they were still carving into stone tablets, and the symbols of the alphabet allowed them to stamp values on coins in a smaller, more condensed version.

The Roman numerical system is still used today although the symbols have changed from time to time. The Romans wrote four as IV, I from V and they wrote six as VI, I to V. Today the Roman numerals are used to represent numerical chapters of books or for the main divisions of outlines. The earliest forms of Roman numeral values are:

Another very common number system we use today is base 2 (binary) which is based on two digits 0 and 1 which can be represented easily using electronic states ( ON and OFF). This is the base of all software and electronic gadgets we enjoy today :)

Last one we discuss here is Hexadecimal number system which is base 16. Now we need 16 symbols. We can take 10 from decimal... What about the other 6... ? First 6 Alphabets are included as symbols (A, B, C, D, E and F) to represent 10 to 15. Hexadecimal is commonly used to represent computer memory addresses.

2AF3 = (2 × 16^{3}) + (10 × 16^{2}) + (15 × 16^{1}) + (3 × 16^{0}) = 10995

Bored of theory and history ? try to solve this famous puzzle to refresh your brain cells.. :)

On the island of Knights and Knaves, every inhabitant is either a knight or a knave. Knights always tell the truth. Knaves never tell the truth; any sentence uttered by a knave is false. A stranger came to the island and encountered three inhabitants, A, B, and C. He asked A, "Are you a knight, or a knave?" A mumbled an answer that the stranger could not understand. The stranger then asked B, "What did he say?" B replied, "A said that there is exactly one knight among us." Then C burst out, "Don't believe B, he is lying!" What are B and C?

**Happy learning :)**

[reference: www.math.wichita.edu]