Why is the last digit of n^5 equal to the last digit of n? [video]

A somewhat more general proof involves the use of the Chinese remainder theorem and the fact that 4 is divisible by the order of the multiplicative group of both integers mod 2 and mod 5. Therefore x^5 = x mod 2 and x^5 = x mod 5 for all x, and as a consequence x^5 = x mod 10 for all x.

This construction seems to fail if the base isn't square free. In particular in base 4 the number 2 has as powers 2, 10, 20, 100 etc.

At work so I haven’t watched the video yet, but my quick rationale is,

Let m = abs(n) It follows n = sign(n) * m We can work with m since sign(n) doesn’t affect the ones digit of n^5. More formally, n^5 = sign(n)^5 * abs(n)^5.

Let m = 10 * b + a, where b = floor(m/10) a = m % 10

Essentially we extract the ones digit as a.

m^5 = (10 * b + a)^5 If we expand the terms, only a^5 affects the ones digit. The remaining has at least one 10 as factor.

Quick python script will verify a^5 has a as the ones digit for 0...9.

  for a in range(10):
      print(pow(a, 5) == a) 

Edit: fixed formatting

good example of an inductive proof, would be helpful for people learning about inductive proofs.