## Powerful Divisibility

#### Problem

Given that n is a positive integer, prove that 21n 5n + 8n is always divisible by 24.

#### Solution

By using the result,
xn yn = (x y)(xn 1 + ... + yn 1)

21n 5n + 8n = (21 5)(21n 1 + ... + 5n 1) + 8n = 16X + 8n 0 mod 8

And by writing the expression differently,
8n 5n + 21n = (8 5)(8n 1 + ... + 5n 1) + 21n = 3Y + 21n 0 mod 3

As the expression is divisible by both 3 and 8, it must be divisible by 24.

Problem ID: 32 (Jan 2001)     Difficulty: 3 Star

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