## Power Divisibility

#### Problem

Consider the following results.

8

8

8

8

8

^{1}1 = 7 = 718

^{2}1 = 63 = 798

^{3}1 = 511 = 7738

^{4}1 = 4095 = 75858

^{5}1 = 32767 = 74681Prove that 8^{n}1 is always divisible by 7.

#### Solution

Clearly it is true for `n`=1: 8^{1}1 = 7.

Assume that it is true for `n`=`k`: 8^{k}1 is divisible by 7.

Consider the next case, `n`=`k`+1.

8^{k+1}1 = 8^{k}81 = 8(8^{k}1)+81 = 8(8^{k}1)+7.

That is, if 8^{k}1 is divisible by 7, 8^{k+1}1 will also divide evenly by 8. As it works for `n`=1, it must be true for all `n`.

Prove that `a`^{n}1 is always divisible by `a`1.

Problem ID: 204 (24 Jan 2005) Difficulty: 3 Star