Quant Boosters  Sibanand Pattnaik  Set 2

Method  1 :
If you are good in binomial then do it directly. Else you can also think this in these lines. Is this not same as 3x + 2y + z = 14 and is this not same as finding the positive distinct integral solutions of the equation a + b + c = 14? so (13c2  3 * 6)/6 = 10Method 2 :
If you want to solve it by binomial then check below
coefficient of x^14 in (x^3 + x^6 + x^9 + ...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)
= coefficient of x^14  6 (1 + x^3 + x^6 + ...) * (1 + x^2 + x^4 + ...) * (1 + x + x^2 + x^3 + ...)
= coefficient of x^8 in (1 + x^2 + x^3 + x^4 + ... + x^8 +...) * (1 + x^2 + x^3 + x^4 + ... + x^8 + ....)
= 1 + 1 +1 + 1 + 1 + 2 + 1 + 2 = 10

Q9) In how many ways can an amount of rs. 100 be paid exactly using coins of denomination 1,2 and 5 such that atleast one coin of each denomination is used?

Basically the problem is to find +ve integer solutions to A + 2B + 5C = 100
So A + 2B can take values 5 , 15, 25 ,......95
The number of solutions corresponding to this will be 2 + 7 + 12 ... + 47 = 245
Also, A + 2B can be 10 , 20 , 30 ... 90
The number of solutions corresponding to this will be 4 + 9 + 14 ... + 44 = 216
so total = 245 + 216 = 461.

Q10) In how many ways can 2310 be expressed as a product of 3 factors?

2310 = 2 * 3 * 5 * 7 * 11
These 5 primes to be distributed into 3 places.
Total : 3^5 ways , but this is ordered ( arranged) and we need unordered.
We know that any triplet with distinct elements (a,b,c) is arranged in 3! ways, so all the distinct triplets in 3^5 = 243 are arranged in 3! Ways.
Except for one triplet (1, 1, 2310) which is arranged in 3!/2= 3 ways
So remove this from total and we are left with only distinct, is divide by 3! ,
So (3^5  3)/3! , and just add that one case of 1, 1, 2310
So (3^53)/3! + 1 = 41

Q11) How many pairs of positive integers (m, n) satisfy 1/m + 4/n = 1/12. Where n is an odd integer less than 60.

(m 12) * (n48) = 12 * 48 = 2^6 * 3^2
now for "n" to be odd , (n48) has to be odd
so n  48 can be either 1 or 3 or 3^2 (because we need ODD)
(m 12) * (n  48) = 2^6 * 3^2 * 1
OR (m 12) * (n  48) = 2^6 * 3 * 3
or (m 12) * (n  48) = 2^6 * 3^2
when n  48 = 1, n = 49 (acceptable)
when n  48 = 3, n = 51 (acceptable)
when n  48 = 3^2, n = 57 (acceptable)
only these 3 values are possible for n < 60 and odd

Q12) Find number of whole number solutions of a + b + c + d = 20 where a, b, c and d ≤ 8

Cofficient of x^20 in (1 + x + x^2 + x^3 + ... + x^8)^4
= coff of x^20 in (1  x^9)^4 * (1  x)^4
= 23c3  4 * 14c3 + 4c2 * 5c3
= 1771  1456 + 60
= 375For more detailed explanation of the concept, refer the video.

Q13) If k is a factor of (n+5) and (6n+54) where n is a natural number, how many possible values of k are there?
a) 6
b) 7
c) 8
d) 9

As given k must be a factor of 6n+30 and is also a factor of 6n+54.
So it must be a factor of (6n+54) – (6n+30) = 24 = 2^3 x 3
Therefore k can take 8 values
Hence (c)

Q14) The sum of all the factors of 480 including 1 and the number itself is 1512. What is the sum of the reciprocals of all the factors?
a) 3.15
b) 2.85
c) 2.55
d) 3.45

Sum of reciprocals = (sum of factors)/480
1512/480= 3.15

Q15) Find the last two digits of 89^102
a) 23
b) 21
c) 01
d) 69

(89^2)^51
89^2 same as 11^2 > 21^51 = 21

Q16) The highest power of 5 in N! is 34. Find the highest power of 7 in N!.
a) 23
b) 26
c) 22
d) Cannot be determined

Highest power of 5 in 125!= 31.
Highest power of 5 in 140!=34
But highest power of 5 in 141!, 142!, 143! And 144! Is also 34.
So N can be 140 to 144, any of these values
However the highest power of 7 in any of these values is same i.e. 22

Q17) A! + B! = N^2, Where A, B and N are whole numbers. How many possible combinations are there?
a) 1
b) 2
c) 3
d) 4

0! + 4! = 5^2
1! + 4! = 5^2
0! + 5! = 11^2
1! + 5! = 11^2
Hence 4 combinations are possible

Q18) x, y and z are natural numbers such that x > y > z > 1 and PRODUCT= 3003. Let A and B be the maximum and minimum values of x + y + z. Find AB.
a) 152
b) 45
c) 107
d) None of the above