Review question

# Can we prove that $26^n$ always ends in the same two digits? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5032

## Question

Prove that, for all positive integral values of $m$, $x-1$ is a factor of $x^m - 1$.

Prove that, if $n$ is an integer greater than $1$, then $26^n - 26$ is an odd multiple of $50$.

Hence, or otherwise, show that, if $26^n$ is evaluated for any integral value of $n$ greater than $1$, then the last two digits of the result are independent of $n$.

[The phrase ‘integral values’ means the same as ‘integer values’. ]