# Quant Boosters - Sibanand Pattnaik - Set 2

• (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

• n^3 means exponent should be a multiple of 3
70= 10 * 7 i.e (9+1) * (6+1) both powers have to be multiple of 3

• Q21) Let D be the number of divisors of a natural number N. Let D1 be the number of divisors of D, D2 the number of divisors of D1 and so on. What will be the final value of Dn?
a) 1
b) 2
c) 3
d) None of the above

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