Powers and remainders

Over on Reddit, a couple of “last digit” puzzles crossed my path, and I thought I’d share the tricks I used, as much for my reference as anything else.

1) Show that the last digit of $6^k$ is 6, for any positive integer $k$.

There’s a standard way to prove this using induction (it’s true for 6, and $6(10k+6) = 60k + 36$, which ends in 6, so every power of 6 must end in 6). However, I prefer a different way:

  • Consider $(5 + 1)^k$
  • The binomial expansion of this is $5^k + \nCr{k}{1} 5^{k-1} + \dots + \nCr{k}{k-1} 5 + 1$
  • Every term except the last is a multiple of 5, so $6^k$ is one more than a multiple of 5.
  • Therefore it ends with 1 or 6.
  • It’s even, so it must end in 6. $\blacksquare$

2) Show that $4^n \equiv 4 \pmod{6}$, for any positive integer $k$.

Again, induction is a lovely way to show this. And again, I’m going to do it differently.

  • Consider $4^n - 4$, which is $2^{2n} - 2^2$
  • Factorise as $\left(2^n - 2\right)\left(2^n + 2\right)$
  • $2^n$ is not a multiple of 3, so either $2^n - 2$ is or $2^n + 2$ is.
  • Therefore $4^n - 4$ is a multiple of 3.
  • It’s also even, so it’s a multiple of 61.
  • So $4^n - 4$ is a multiple of 6, which means $4^n \equiv 4 \pmod{6}$.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. and, for that matter, 12 []


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