TIFR CSE 2018 | Part B | Question: 1

Question

What is the remainder when $4444^{4444}$ is divided by $9?$

• A. $1$
• B. $2$
• C. $5$
• D. $7$
• E. $8$

Answer 1

see that 4444 (mod 9) = 7

4444^4444 (mod 9)  = 7^4444 (mod 9) = (-2)^4444 (mod 9) = 16^1111 (mod 9) = 7^1111 (mod 9) = [ 7*49^555 ] (mod 9) = 7 (mod 9) * 49^555 (mod 9) = 7 *   4^555 (mod 9) = 7* 64^185 (mod 9) = 7 *  1^185 (mod 9) = 7

 

using formulas:

a^k (mod b) = (a mod b)^k (mod b)

Answer 2

4444^4444  mod 9=(7^4444) mod 9 now we will see the remainders which power of 7 gives while dividing with 9

7^1mod9=7

7^2mod9=4

7^3mod9=1 so remainder will repeat after this

now divide 4444 with 3 the remainder which it gives that term is the final remainder (term refers here to remainder whch we got while dividing powers of 7)

so 4444 mod 3=1 so 1st term remainder which is 7 earlier

Answer 3

USING MADE EASY APTITUDE

Answer 4

7 should be remainder