Sols to Assignment 24
62. "Last two digits" means "mod 100." We have φ(100) = 40. Since 403 ≡ 3
(mod 40), Corollary 4.32 implies that 123403 ≡ 1233 ≡ 233 (mod 100).
The last congruence is because 123 ≡ 23 (mod 100). A short calculation
yields 233 = 12167, so the answer is 67.
66. (a) We have 34 = 81 ≡ 1 (mod 10) [which is also evident from Euler's
Theorem since φ(10) = 4] , so we have to look at only 31 ≡ 3, 32 ≡ 9,
33 ≡ 7, and 34 ≡ 1 (mod 10). The possible last digits are 1, 3, 7, 9.
(b) We can't use Euler's theorem for 10 because gcd(4, 10) ≠ 1. However,
we can simply compute 41 ≡ 4, 42 ≡ 6, 43 ≡ 4, …, to see that the possible
last digits are 4 or 6.
(c) The powers of 6 are 6, 36, 216,…, so the only possible last digit is 6.
(d) Since φ(10) = 4, Euler's Theorem tells us that 74 ≡ 1 (mod 10). The powers
of 7 mod 10 are 7, 72 ≡ 9, 73 = 343 ≡ 3, and 74 = 2401 ≡ 1 (mod 10). Therefore,
the possible last digits are 1, 3, 7, 9.