Quant Boosters  Hemant Yadav  Set 1

From the 4th point, m = 2n/10
As 'n' has 'k' digits, 2n will have either 'k' digits or 'k + 1' digits. Now 'm' has got to have 'k' odd digits, hence 2n must have 'k + 1' digits with the last digit obviously 0 (as 'n' is divisible by 5) and all the remaining digits odd. So, 'm' is also a 'k' digit number and combining that with the 4th point we can say that 'm' is a 'k' digit number with all of its digits odd.
Hence, 'n' must have its digits such that each digit of 'n' when multiplied by 2 gives a carry of 1, if this doesn't hold then we will get an even digit in 'm', which is not acceptable.
As each digit of 'n' when multiplied by 2, gives a carry forward of 1, each digit of 'n' should be greater than equal of 5 with the last digit always 5 (as 'n' is a multiple of 5)
For k=1, possible cases =1
For k=2, last digit can be selected in 1 way and 2nd last digit can be selected in 3 ways ( 5 or 7 or 9)
For k=3, last digit in 1 way, other 2 digits in 3*3 ways
Generalizing, 'n' can be formed in 3^(k  1) ways

Q11) For how many natural numbers 'p', the three numbers (p  1)/4, (p + 1)/2 and p are always prime?
[OA : 1]

Say, its 6n + 1, then 2k + 1 = 12n + 3, so a multiple of 3. If its 6n  1, then 4k + 1 = 24n  3, so a multiple of 3. Hence, k is either 2 or 3. Only 3 satisfys. So, p = 13 is the only value for which all three numbers are prime.

Q12) "A" and "B" are playing series of repeated games. A player wins if he wins 2 games before other player. Probabilities of "A" winning, drawing and loosing a game are 1/2, 1/3 and 1/6. What is the probability of "A" winning the series.
[OA : 27/32]

If P(n) denotes the probability of winning of A when no of wins of B is n, then
P(0) = (1/2)² + 2(1/3)(1/2)² + 3(1/3)²(1/2)² + ....
= (1/2)²{1 + 2/3 + 3/9 + 4/27 + ....)S = 1 + 2/3 + 3/9 + 4/27 + ...
S/3 = 1/3 + 2/9 + 3/27 + ....
=> 2S/3 = 1 + 1/3 + 1/9 + 1/27 + ....
=> S = 9/4
P(0) = (1/4)(9/4) = 9/16P(1) = (2!/(1!*1!))(1/6)(1/2)² + (3!/1!*1!*1!))(1/3)(1/6)(1/2)² + (4!/(2!*1!*1!)(1/3)²(1/6)(1/2)² + ....
= (1/6)(1/2)²{2 + 6(1/3) + 12(1/3)² + 20(1/3)³ + 30(1/3)^4 + ... )S = 2 + 6(1/3) + 12(1/3)² + 20(1/3)³ + 30(1/3)^4 + ... .....(1)
=> S/3 = 2/3 + 6(1/3)² + 12(1/3)³ + .......... (2)
(1)  (2) => 2S/3 = 2 + 4/3 + 6/3² + 8/3³ + .... ............(3)
=> 2S/9 = 2/3 + 4/3² + 6/3³ + ..... ...........(4)
(3)  (4) => 4S/9 = 2 + 2/3 + 2/3² + 2/3³ + .... = 3P(1) = (1/6)(1/2)²(27/4)
= 9/32=> Probability of A winning = 9/16 + 9/32 = 27/32

Q13) Find the smallest n for which rightmost nonzero digit of P(n) will be odd where P(n) = (n + 5)!/(n – 3)!

p(n) = (n + 5)!/(n  3)!
It is clear that p(n) is a product of 8 consecutive numbers and divisible by atleast 8!. That means minimum power of 2 in p(n) is 7, i.e, p(n) is divisible by 2^7.
But last nonzero digit has to be an odd number, so p(n) should be divisible by 5^7.
So 15625 (5^6) or its multiple has to be one of the consective numbers, and we have to make sure that there is no multiple of 16 among the 8 numbers.
Multiple of 16 nearest to 15625 are 15616 and 15632.So, least value of n = 15620.
p(n) = 15618 * 15619 * 15620 * 15621 * 15622 * 15623 * 15624 * 15625
After removing (2^7)*(5^7), we will get
7809 * 15619 * 781 * 15621 * 7811 * 15623 * 1953 * 1Unit digit = unit digit of 9 * 9 * 1 * 1 * 1 * 3 * 3 * 1 = 9

Q14) How many real number solutions are there for a, b, c and d such that a² + b² + c² + d² = a(b + c + d)
[OA : Only one solution]

a² + b² + c² + d² = a(b + c + d)
=> a²/4 + (a²/4 + b²  ab) + (a²/4 + c²  ac) + (a²/4 + d²  ad) = 0
=> a²/4 + (a/2  b)² + (a/2  c)² + (a/2  d)² = 0
=> a = b = c = d = 0 is the only solution

Q15) There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner. What is the probability that they don't collide?
[OA : 1/4]

Each ants can go towards two direction. so total different possibilities are 2 * 2 * 2 = 8. Out of which 2 are favorable where they all goes clockwise or anticlockwise . Hence required probability = 1/4

Q16) A motor cyclist is travelling on a straight road at 70 kmph towards his house. His pet bird travels at 80 kmph from the motorcylist to the house and back in 1.6 hours. Find the distance between the house and the place from where the bird starts its journey towards the house?
[OA : 120 km]

In 1.6 hours the bird flies 128 km and in the same time the motor cyclist goes 112 km and they travel a distance twice the the distance of the house. So distance is (128 + 112)/2 = 120 km

Let x = a, y + z = b
we get (x + y + z)^2006 + (x  y  z)^2006 = k1 * a^2006 * b^0 + k2 * a^2004 * b^2 + .... + k1004 * b^2006
=> number of terms = sum of the number of terms in b^0, b^2, .... , b^2006
= 1 + 3 + 5 + ... + 2007
= 1004² = 1008016

a) boxes and balls are all distinct and only 1 ball in one box
Its like permuting a 5 digit number, so 5! Waysb) boxes are same and balls are distict and only 1 ball per box
Since all the boxes are same and one ball per box, so just one wayc) balls are same and boxes are different and only 1 ball per box
Here all the balls are same and every box has one ball, so again one wayd) balls and boxes are same and only 1 ball per box
Here also only only waye) repeat all above questions with any number of balls in any box
i) 5^5 ways, as all the balls and all the boxes are different.
ii) Here boxes are same but balls are different.
(5, 0, 0, 0, 0) – 1 way
(4, 1, 0, 0, 0) – C(5, 4) = 5 ways
(3, 2, 0, 0, 0) – C(5, 3) = 10 ways
(3, 1, 1, 0, 0) – C(5, 3) = 10 ways, we don’t have to choose one of other two as they will automatically choose any of the two boxes (all the boxes are similar)
(2, 2, 1, 0, 0) – C(5, 2)*C(3, 2)/2 = 15, here we need to divide by 2 as as it doesn’t matter if I put the first pair in box 1 and second pair in box 2 or viceversa
(2, 1, 1, 1, 0) – C(5, 2) = 10 ways
(1, 1, 1, 1, 1) – 1 way
So, total = 1 + 5 + 10 + 10 + 15 + 10 + 1 = 52 waysiii) x1 + x2 + x3 + x4 + x5 = 5
So, C(9, 5) = 126 waysiv) Since all balls and boxes are same, we just need to find what different combinations are possible:
(5, 0, 0, 0, 0)
(4, 1, 0, 0, 0)
(3, 2, 0, 0, 0)
(3, 1, 1, 0, 0)
(2, 2, 1, 0, 0)
(2, 1, 1, 1, 0)
(1, 1, 1, 1, 1)
So, 7 ways

Q17) Find the largest possible value of k for which 3^11 is expressible as the sum of k consecutive positive integers. What will be the answer if 3^11 is expressed as sum of k consecutive integers.
[OA : 486, 2 * 3^11]

Let the first of the consecutive integers be a + 1, then the consecutive integers will be a + 1, a + 2, ..., a + k
So, (a + 1) + (a + 2) + .... + (a + k) = 3^11
k(k + 2a + 1) = 2 * 3^11Now for the first part all the integers are positive, so both (k + 2a + 1) > k
Hence maximum value of k will be 2 * 3^5 = 486For the second part there is no such constraint, so maximum possible value of k will be 2 * 3^11

Q18) Find the sum of 6 * 1! + 13 * 2! + 22 * 3! + 33 * 4! + 46 * 5! + ... n terms

It is (n +2)! + 2[(n + 1)!]  4 = (n + 4)[(n + 1)!]  4.
Key lies in simply finding the nth term of the sequence which is (n + 1)[(n + 1)!] + 2n[n!].

Q19) How many natural numbers N < 40000 such that cube of sum of digits of N gives back the same number N.
For eg, 8³ = 512, sum of digits is 8, same as cube root of 512.[OA : 6]