mathschallenge.net

Problem

Consider the following results.

99¹ = 99
99² = 9801
99³ = 970299
99⁴ = 96059601
99⁵ = 9509900499

Prove that 99^$n$ ends in 99 for odd $n$.

Solution

Given that $a$ $b$ mod $k$, $a$^$n$ $b$^$n$ mod $k$.

As 99 1 mod 100, 99^$n$ (1)^$n$ mod 100

For odd $n$, (1)^$n$ = 1.

Therefore 99^$n$ 1 mod 100 for odd $n$; that is, one less than a multiple of 100, which means it will end in 99.

Prove that, for odd $n$, (i) 9^$n$ ends in 9, (ii) 999^$n$ ends in 999, (iii) Generalise.

If $a$ $b$ mod $k$, then prove that $a$^$n$ $b$^$n$ mod $k$.