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

rejected and -2 answer

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

Let 3 statements be made

(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