## Finishing With 99

#### Problem

Consider the following results.

99^{1} = **99**

99^{2} = 9801

99^{3} = 9702**99**

99^{4} = 96059601

99^{5} = 95099004**99**

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`.

Problem ID: 141 (Dec 2003) Difficulty: 4 Star