Quant Boosters  Hemant Malhotra  Set 10

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 = 4or direct 6k1=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
 10^2p  10^p + 1 is divisible by 13 for the largest integer p < 10
 The remainder on dividing 16! + 89 by 323 is q
 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 = 7N = 16! + 89 mod 323
323 = 17 * 19
N mod 17 = 1 + 4 = 3
N mod 19 = 9 + 13 = 3
q = 346c23 mod 23 = = 2=r
p + q + r = 12