# Quant Boosters - Sibanand Pattnaik - Set 2

• 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

• You will finally end up with a prime number which will have 2 divisors. 2 has 2 divisors. Hence (b)

• Q22) In how many ways can 3660 be written as the sum of 2 or more consecutive positive integers?
a) 2
b) 3
c) 4
d) More than 4

• Direct formula for natural numbers = number of odd factors - 1
So 3660 = 2^2 * 3 * 5 * 61
so odd factors = 8
Answer = 8 - 1 = 7
as in case of "integers" its 2 * (number of odd factors) - 1

Else you can think this way :

3660 = 61 x 3 x4 x5
3660= a + (a+1) + (a+2)...+ (a+(n-1)
Then 3660 = n x Average of the above series
If the number of terms in the series is odd, then the middle term of the series will be the average
So say the number of terms is 61, then the middle term will be 3x4x5= 60. So the series will be: (30+31+32...60+61+62...90)
If the number of terms is 3, the average will be 61x4x5= 1220. So the series is 1219+1220+1221.
Therefore if the number of terms is odd, its easy to construct a series like this
3660 has 7 odd factors other than 1

• Q23) N has 6 factors and lies between 250 and 350. How many values of N are possible?
a) 12
b) 13
c) 14
d) None of the above

• If a and b are prime numbers then N= a^5 or N= a^2 x b
N= a^5 :No values
N= a^2 x b
If a= 2, then b can take prime values of 67, 71, 73, 79, 83: 5 values
If a= 3, then b can take prime values of 29, 31, 37: 3 values
If a= 5, then b can take prime values of 11, 13: 2 values
Total values: 10

• Q24) N^2 has 15 factors. M^2 has 9 factors. Which of the following cannot be the number of factors that MxN can have, if M and N are co prime to each other?
a) 24
b) 32
c) 20
d) 40

61

1

61

63

63

61

62

61