# 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 * 102 + 2 * 101 + 3 * 100

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 * 102+ 101* 2 + 100* 3 + 1 * 10-1+ 10-2* 2 + 10-3* 3

Base System

When we wrote123 = 1 * 102+ 101* 2 + 100* 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)digit1+ 10(n-2)digit2+… + digitn (Digits taken left to right)

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

(a + b)2= a2+ 2ab + b2

(a - b)2= a2- 2ab + b2

(a + b)(a - b) = a2- b2

a3+ b3= (a + b) (a2- ab + b2)

a3- b3= (a - b) (a2+ ab + b2)

(a + b)3= a3+ 3a2b+ 3ab2+ b3

(a - b)3= a3- 3a2b+ 3ab2- b3

(a + b + c)2= a2+ b2+ c2+ 2ab + 2bc + 2ca

(a + b + c) (a2+ b2+ c2- ab - bc - ca) = a3+ b3+ c3- 3abc

a3+ b3+ c3= 3abc, if a + b + c = 0

a0= 1

a1= a

am× an= a (m + n)

am/ an= a ( m – n )

(am)n= am.n

a- m= 1 / am

(ab)m= ambm

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 2nif the last n digits are divisible by 2n.

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!  :)

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