Arrange the following in ascending order : (A) Remainder of 4916 when divided…

Solution:

• Remainder of 49^16 when divided by 17: 49 ≡ 15 (mod 17), and by Fermat's little theorem 15^16 ≡ 1 (mod 17). Therefore the remainder is 1.

• Remainder of 155^17 when divided by 17: 155 ≡ 2 (mod 17), and 2^17 = 2*(2^16) ≡ 2*1 ≡ 2 (mod 17). Therefore the remainder is 2.

• Remainder of 2^446 when divided by 9: The powers of 2 modulo 9 have period 6 (Euler's theorem). Since 446 ≡ 2 (mod 6), 2^446 ≡ 2^2 ≡ 4 (mod 9). Therefore the remainder is 4.

• Last digit of 7^45: Last digits of powers of 7 cycle every 4: 7, 9, 3, 1. Since 45 ≡ 1 (mod 4), the last digit is 7.

Thus the numeric values are: remainder of 49^16 mod 17 = 1; remainder of 155^17 mod 17 = 2; remainder of 2^446 mod 9 = 4; last digit of 7^45 = 7.

Ascending order by value: remainder of 49^16 when divided by 17 (1), remainder of 155^17 when divided by 17 (2), remainder of 2^446 when divided by 9 (4), last digit of 7^45 (7).