mathschallenge.net logo

Power Divisibility

Problem

Consider the following results.

81minus1 = 7 = 7times1
82minus1 = 63 = 7times9
83minus1 = 511 = 7times73
84minus1 = 4095 = 7times585
85minus1 = 32767 = 7times4681

Prove that 8nminus1 is always divisible by 7.


Solution

Clearly it is true for n=1: 81minus1 = 7.

Assume that it is true for n=k: 8kminus1 is divisible by 7.

Consider the next case, n=k+1.

8k+1minus1 = 8k8minus1 = 8(8kminus1)+8minus1 = 8(8kminus1)+7.

That is, if 8kminus1 is divisible by 7, 8k+1minus1 will also divide evenly by 8. As it works for n=1, it must be true for all n.

Prove that anminus1 is always divisible by aminus1.

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

Only Show Problem