# Quant Boosters - Sibanand Pattnaik - Set 2

• 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 (3^5-3)/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) * (n-48) = 12 * 48 = 2^6 * 3^2
now for "n" to be odd , (n-48) 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
= 375

For 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 A-B.
a) 152
b) 45
c) 107
d) None of the above

• 3003= 3 x 7 x 11 x 13
A= 143+7+3= 153
B= 21 + 11 + 13= 45
A-B= 108
Hence (d)

• Q19) How many natural numbers less than 4444 are "not co primes" to 247?
a) 585
b) 341
c) 244
d) 557

• 247= 13 x 19
So all multiples of 13 and 19 has to be excluded till 4444
Multiples of 13 : 341
Multiples of 19: 233
Multiple of both 13 and 19 i.e. of 247: 17
Therefore number of numbers less than 4444 which are co-prime to 247: 341+233-17= 557

• Q20) N , a natural number has 2 prime factors. N^3 can have which of the following number of factors?
a) 18
b) 90
c) 70
d) 84

61

61

61

64

58

61

61

62