# Quant Boosters - Hemant Malhotra - Set 10

• 82! mod 83 = -1
similarly 40!^2 * 41 * (-41)mod 83=-1
40! * 41 mod 83=1
if 40! mod 83=2
then 2 * 41 mod 83 = -1
but -2 * 41mod 83 = -82 mod 83 = 1
so it's -2 = 81

• Q26) Find remainder when 19! is divided by number of digits in 23!

• 21!, 22! ,23! and 24! have same digits so 19! mod 23 = 4

19! mod 23=4
21! mod 23=1
so 21 * 20 * 19! mod 23 =1
-3 * -2 * 19! mod 23=1
6 * 19! mod 23=1
6r = 23k + 1
so r = 4

or direct 6k-1=23 so k=4

• Q27) Find remainder when (105)^13! is divided by 103

• E(103)=102
105^102 mod 103=1
so now we have to check 13! will leave what value of remainder with 102
13! mod 102
102 = 6 * 17
13! mod 6=0
13! mod 17= 3
6a = 17b + 3 so 54
2^54 mod 103 = 8

• Q28) What is the remainder when ((55)^15!)^188 is divided by 17?

• 15! is multiple of 16 and E(17)=,16 so 55^16k mod 17=1

2nd approach - 55 mod 17=4
(4^15!)^188 mod 17
(16)^15!/2 )^188 mod 17
(-1)^even mod 17 = 1

• Q29) What would be remainder when 8! is divided by 17

• 2^7 * 3^2 * 5 * 7 mod 17
(-8) * (-8) * 35 mod 17
64 * 35 mod 17
13

• Q30) Let 3 statements be made

1. 10^2p - 10^p + 1 is divisible by 13 for the largest integer p < 10
2. The remainder on dividing 16! + 89 by 323 is q
3. 46C23 leaves remainder r on division by 23

Then p + q + r equals

(a) 12
(b) 19
(c) 26
(d) 33
(e) none of the foregoing

• a^2 - a + 1 mod 13 = 0
a = 10^p
a(max) = 10^9
10^18 - 10^9 + 1 mod 13 = 1 + 1 + 1 = 3
10^16 - 10^8 + 1 mod 13 = 3 - 9 + 1 = 8
10^14 - 10^7 + 1 mod 13 = 9 + 3 + 1 = 13 mod13=0 so
p = 7

N = 16! + 89 mod 323
323 = 17 * 19
N mod 17 = -1 + 4 = 3
N mod 19 = 9 + 13 = 3
q = 3

46c23 mod 23 = = 2=r

p + q + r = 12

68

61

107

34

61

34

61

61