## Cubes And Multiples Of 7

#### Problem

Prove that for any number that is not a multiple of seven, then its cube will be one more or one less than a multiple of 7.

#### Solution

Given `n` is any number that is not evenly divisible by 7, let `n` = 7`a` + `b`, where `b` = 1,2,3,4,5,6.

`n`^{3} = (7`a` + `b`)(7`a` + `b`)(7`a` + `b`)

Clearly the only term under expansion that does not have at least one multiple of 7 will be `b`^{3} and as `b` = 1,2,3,4,5,6 we get `b`^{3} = 1,8,27,64,125,216.

`b`^{3} 1,1,-1,1,-1,-1 mod 7, respectively.

Hence `n`^{3} 1 mod 7, where `n` is not a multiple of 7.

Corollary

`n`^{3} -1,0,1 mod 7, where `n`**N**.

Problem ID: 24 (Nov 2000) Difficulty: 3 Star