Remainder Theorem Practice Questions (Solved) - Hemant Malhotra

• Find remainder when 6^83 + 8^83 divided by 49

E(49)=49 * 6/7=42
so 6^42 mod 49=1
and 8^42 mod 49=1
now 6^83=6^84 * (1/6)
and 8^(83)=8^84/8
so 1/6+1/8 mod 49
14*(48^-1) mod 49
now 48 mod 49=-1
so 14 * (-1)^-1 mod 49
-14 mod 49
so 49 - 14=35

OR if you want a much faster method,
(7-1)^83+(7+1)^83
expand this binomial expension
even powers will get cancelled only odd will be there and 2 times
so 2(7^83 + 83c1 * 7^81.........83c8227) mod 49
now here all terms are div by 49 only 2 * 83 * 7 will not be div by 49
so 2 * 83 * 7 mod 49=35

3^33^333^3333^....^3333...(100 digits) mod 455 = ?

3^33^333^3333.........100 digts mod 5
E(5)=4
so 33^333^333.^..100 digits mod 4=1
so 3^1 mod 5=3
now 3^33^333^333 .... mod 7
E(7)=6
33^333^3333^.... mod 6
6 = 2 * 3
33^333^333^333... mod 2=1
33^333^3333... mod 3=0
2a+1=3b
so 3
so 3^3 mod 7 =27 mod 7=6
now for 13
E(13)=12
33^333^3333^... mod 12
12=3*4
33^333^3333.. mod 3=0
and mod 4=1
so 3a=4b+1 so 9 so 3^9 mod 13 so 27^3 mod 13=1
so 5x+3=7y+6=13z+1 so 118

Find Remainder when 2^100+ 2^200 + 2^300 +.....+ 2^100000 divided by 7

2^100+4^100+8^100+16^100............2^1000)^100 mod 7
now 2^100 mod 7=2
4^100 mod 7=4
and 8^100 mod 7=1
now remainder in these three cases =2+4+1=7 mod 7=0
so cycle will repeat like this in 3
so total terms =
2^1000=2*(2)^(n-1)
2^1000=2^n
n = 1000 terms
so 1000 mod 3=1
so only last term will count so remainder is 2

if N=2222.....123 times then find Remainder when N is divided by 41

Make group of 5 from right and add them if sum is div by 41 then num is div so 24 * 22222 + 222 mod 41 = 17
(1 * 271 * 9 = 99999 = 10^5 - 1 that's y we make group of 5)

Find remainder when 40! mod 83?

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=-82mod 83=1
so it's -2=81

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

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

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

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

(P) 10^2p - 10^p + 1 is divisible by 13 for the largest integer p < 10
(Q) The remainder on dividing 16! + 89 by 323 is q
(R) 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

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