# Quant with Kamal Lohia - Part 14

• Q1) How many factors of 2010^2010 have last digit 2?

2010^2010 ≡ (2×3×5×67)^2010.
Now see that unit digits of positive powers of 2 are: 2, 4, 8, 6 as powers are of the form 4k + 1, 4k + 2, 4k + 3 and 4k respectively.
Also with similar pattern in powers, unit digits of 3 are: 3, 9, 7, 1 and that of 7 are: 7, 9, 3, 1.
Unit digit of positive powers of 5 is always 5, so we need to consider the divisors which contain 50 = 1 only.
Now you just need to make the cases: i.e. if powers of 2 end in 2 {i.e. in 503 ways}, then powers of 3 and 7 can end in (1, 1), (3, 7), (7, 3) and (9, 9) {i.e. in 502² + 503² + 502² + 503² ways}. So total cases - 2×503(502² + 503²).
If powers of 2 end in 4 {i.e. in 503 ways}, then powers of 3 and 7 can end in (3, 1), (1, 3), (9, 7) and (7, 9) {i.e. 503×502 + 502² + 503² + 502×503 = 1005² ways}. So total cases - 503 × 1005².
Similarly solve for other two cases

Q2) Answer the questions on the basis of the information given below.
Several teams participated in a football tournament. No match played in the tournament was abandoned and all the matches either resulted in a win/loss or a draw. The matches played in the tournament had one or the other of the score lines (1 - 0), (2 - 1), (2 - 0), (3 - 1), (5 - 1), (4 - 1), (5 - 0), (0 - 0), (1 - 1), (2 - 2) and (3 - 3). The following table provides information about the total number of goals in a match and correspondingly the %age of matches in the tournament with the same number of goals.

The number of matches that resulted in a draw is 22 percentage points less than the percentage of matches that resulted in a win/loss.

(1) Out of the matches in which a total of 2 goals were scored, what could be the minimum number of wins/losses as a percentage of the total number of such matches?
(2) Let the total number of matches in the tournament be 200. No match in which 6 goals were scored ended in a draw. Then the number of matches that resulted in a win/loss and a total of 2 goals does not exceed
(3) The total number of matches in which only one team scored goal(s) as a percentage of the total number of matches played in the tournament cannot be more than

First adding all the given percentage of matches, we get to know that 10% of matches are there where no goal was scored. i.e. If I assume there are total 100 matches, then we can frame a table of matches with their respective goals as follows:

Goals -> Matches
0 -> 10
1 -> 5
2 -> 36
3 -> 18
4 -> 15
5 -> 8
6 -> 8

Now total number of draw matches is 39% from the given data. Matches with 0 goals (i.e. 10 matches) are certainly drawn. Remaining 29 drawn matches must be from 2 goals, 4 goals or 6 goals only.
(1) For minimum matches with 2 goals having win/loss, we should have all the 29 drawn matches in 2 goals category only. So out of 36 matches with 2 goals, 29 are drawn and remaining 7 matches result in win/loss.
So the required percentage is (7/36)*100 = (175/9)%
(2) Now total number of matches has been given to 200. So all previously assumed number of matches will be doubled. Also mentioned that no 6 goal match was a draw. So all 58 drawn matches should be from 2 goal matches (i.e.72 matches) 4 goal matches (i.e. 30 matches) only.
For maximum value of 2 goal matches resulting in win/loss we must accommodate maximum number of drawn matchs in 4 goal ones. So atleast 28 2-goal matches has to be drawn i.e. maximum 72 - 28 = 44 2-goal matches result in win/loss.
(3) Now we need to find maximum number of matches in which only one team is scoring goals. From the data given for score line, we can see that only in 1-goal, 2-goal and 5-goal matches this feat is possible. So maximum (5 + 36 + 8 ) = 49% matches can exhibit the desired score line.

Q3) What is the remainder when 123456789 is divided by 1001

1000 = -1 mod1001
1000² = 1 mod1001
1000³ = -1 mod1001
So If N = 123 456 789 = 123 * 1000² + 456 * 1000 + 789
then N = {123(1) + 456(-1) + 789} mod1001 = {123 - 456 + 789} mod1001 = 456 mod1001.
So basically rule becomes that make the groups of three consecutive digits starting from right most digit of the number and the alternate groups are positive and negative with first rightmost group being positive.
That means if my numbers is 111111 then there will be two groups of three digits and the right most digit group is positive then next one negative i.e. -111 + 111 = 0 mod1001.
Also if you consider a four digit number say 1234, it will have two groups as -1 + 234 i.e. 233 mod1001.
So you can go on repeating this procedure but with smaller numbers only.

Q4) How many distinct triangles of integer sides are possible having perimeter as 20cm?

No side of triangle can be 10 as then sum of other two side cannot be greater than 10. So maximum side length possible for any side is 9.
Now using reverse approach, we can distribute 9 to each of the sides so that we need to remove the extra 7 now (9 * 3 - 20 = 7) from the three sides i.e. whole number solution of the equation a + b + c = 7 i.e. C(9, 2) = 36.
But this includes the ordered pairs of solution which needs to be unordered as follows:
There are 'x' number of triplets which are counted 3! = 6 times where all the three numbers are different (e.g. 1,2,4)
There are 'y' number of triplets where two of the three numbers are same (e.g. 1,1,5) and are counted 3!/2! = 3 times. Here y = 4 (because the identical two numbers can be 00, 11, 22 or 33 only)
There are 'z' number of triplets where the three numbers are same and are counted once only. Here z = 0 (because 7 is not divisible by 3)
So equating the total ordered triplets, we have 6x + 3y + z = 36.
That means 6x + 3(4) + (0) = 36
=> x = 4 and total unordered triplets are: x + y + z = 4 + 4 + 0 = 8.

Using forward approach we can use the condition that maximum length of any side is 9, now writing all lengths in order we have:

1. 992
2. 983
3. 974
4. 965
5. 884
6. 875
7. 866
8. 776

These are the only eight triangles possible integral side lengths and perimeter 20.

Q5) In how many ways can a sum of 12 be achieved If 3 dices are thrown

You can use forward or reverse approach but shouldn't neglect the conditions on A, B, C. Here A, B, C are natural numbers varying from 1 to 6.
FORWARD approach is to distribute 12 among A, B, C following the above conditions i.e. natural number solutions of A + B + C = 12 such that A, B, C < 7.
Now without upper restriction, number of natural number solutions of this equation = C(11, 2) = 55.
[Hint: Whole number solutions of A + B + C + ..... r terms = n is given by C(n + r - 1, r - 1)]
This include the cases where the restriction is violated. So let's calculate the cases where the condition is being violated for sure i.e. one of A, B, C is 7 (which can be selected in 3 ways) and remaining two are natural numbers.
These solutions number to = 3 * C(5, 2) = 30.
Hence the required number of solutions are 55 - 30 = 25.

REVERSE approach is that give 6 to each of the three A, B, C. So their sum becomes 18. Now we need to remove 6 from the sum with the restriction that complete 6 shouldn't be removed from any one of A, B or C (which can be done in 3 ways only).
So total number of ways to remove 6 from the sum of A, B and C i.e. whole number solutions of A' + B' + C' = 6 is given by = C(8, 2) = 28.
And after subtracting the restricted 3 cases (as discussed in above para), we are left with the required number of solutions as = 28 - 3 = 25.

Q6) If N is the number of ways of dividing 500 students into 5 distinctly identifiable sections of 100 students each, how many zeroes does N ends with ?
a. 0
b. 2
c. 5
d. 4

First you are selecting 100 from 500 in C(500, 100) = 500!/(100!)(400!) ways.
Then you are selecting next 100 from remaining 400 in C(400, 100) = 400!/(100!)(300!) ways.
Then you are selecting next 100 from remaining 300 in C(300, 100) = 300!/(100!)(200!) ways.
And last you are selecting 100 from remaining 200 in C(200, 100) = 200!/(100!)(100!) ways.
So total number of ways comes out to be 500!/(100!)^5
500! ends with 124 zeroes and 100!ends with 24 zeroes.(100!)^5 ends with 120 zeroes.
hence 124-120= 4 zeroes

Q7) There are 35 students in a hostel. If the number of students increases by 7, the expenditure of mess increase by Rs. 42 per day while the average expenditure per head diminishes by Re. 1. Find the original expenses of the mess.

First-normal math way - Let x be the original daily per head expenditure. So equating daily expense calculated in two ways, we get: 35x + 42 = 42(x - 1)
=> x = 12.

Second - Just look for the difference in expense per head. Due to addition of 7 students, expense of Rs 42 has been added and expense per person is decreased by Re 1. So total expense for the 7 new students = 35 + 42 = 77 i.e. 11 per head.
=> Original expense per person is 11 + 1 = 12.

Q8) Two persons,X and Y,start together from the starting point and run on a 4 km track in a race of 16 km.The ratio of their speeds is 3:7. How often do they meet on the track in the race if they both run in the clockwise direction?
a) once
b) twice
c) thrice
d) five times

It's a 4 km track and the race is for 16 km. That means winner will be completing 4 complete rounds for sure. Right?
As the speed of two persons is in the ratio 3 : 7, so faster one is completing 4 rounds while slower one will complete (3/7)*4 = 1 + (5/7) rounds. Is that ok?
Now observe carefully. The two runners are moving in same direction from same point and starting simultaneously. So distance between them is increasing continuously and when it will become 1 complete round they will be meeting for first time. At this time the difference between the distance travelled by the two runners is 1 complete round.
Continuing there journey, difference of distance travelled by two runner will start increasing further. But they will meet for the second time only when that difference of distance travelled becomes 2 complete rounds.
And we can take a note from here that they will not meet third time till the differnce in distance travelled is less than 3.
In this case, the difference in distance travelled is = 4 - 12/7 = 16/7 < 3. So they will be meeting 2 times while moving in same direction.

Q9) A faulty odometer of a car always jumps from digit 4 to digit 6, always skipping the digit 5, regardless of the position. For example, after travelling for one kilometer the odometer reading changed from 000149 to 000160. If the odometer showed 000000 when the car was bought and now it shows 003008, how many kilometers has the car travelled?

Now final reading shown is 3008 and the initial reading was 0000. Also the meter skips the digit 5 at every place. So we just need to count the number of numbers from 1 to 3008 which do not contain 5.
There can be many ways but one of the fastest can this one:
After 3000 only 7 values are possible (except 3005). Now we need to count the favourable numbers from 1 to 3000 which is equivalent to the number in the range 0000 to 2999. Now we need to make four digit numbers where first digit can be filled in 3 ways (i.e. 0, 1 or 2) and all other three digits can be filled in 9 ways each (except 5). So total such 4-digit numbers are 3 * 9 * 9 * 9 = 2187.
Adding we get 2187 + 7 = 2194 as the final answer.

Q10) 4 balls are to be selected from a group of 11 balls. 5 of them are of type A, 2 of them are type B, 2 are of type C, one of type D and one of type E.
(1) In how many ways we can select 4 balls out of these 11 balls?
(2) In how many ways we can arrange 4 balls out of these 11 balls in a line?

(1) It is the coefficient of x^4 in the expansion of (1 + x)^2 (1 + x + x^2)^2 (1 + x + x^2 + x^3 + x^4 + x^5) which is equal to 31.
OR it is the whole number solution of A + B + C + D + E = 4 where A, B, C, D, E are whole numbers such that A ≤ 5, B ≤ 2, C ≤ 2, D ≤ 1, and E ≤ 1 which is given by = C(8, 4) - 2C(5, 4) - 2C(6, 4) + 1 = 31

You could have also calculated it by simple case works also.
You want to select 4 out of 5A, 2B, 2C, 1D, 1E - so there are five cases:
(i) All four same - i.e. only 1 case.
(ii) Three same and 1 different - 1 * 4 = 4 cases.
(iii) Two same and Two same - C(3,2) = 3 cases.
(iv) Two same and 1 different and 1 different - 3 * C(4,2) = 18 cases.
(v) All four different = C(5,4) = 5 cases.
Total selections possible = 1 + 4 + 3 + 18 + 5 = 31.

(2) Now we can easily find the number of arrangements in each of the five cases individually and add them to get the total arrangements possible.
(i) = 1C(4, 4) = 1
(ii) = 4C(4, 1) = 16
(iii) = 3C(4, 2) = 18
(iv) = 18C(4, 2)C(2, 1) = 216
(v) = 5 * 4! = 120
Total arrangements possible = 1 + 16 + 18 + 216 + 120 = 371.

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