If you raise any integer to the fifth power, its last digit doesn’t change. For example, 2^{5} = 32.

It’s easy to prove this assertion by brute force. Since the last digit of *b*^{n} only depends on the last digit of *b*, it’s enough to verify that the statement above holds for 0, 1, 2, …, 9.

Although that’s a valid proof, it’s not very satisfying. Why fifth powers? We’re implicitly working in base 10, as humans usually do, so what is the relation between 5 and 10 in this context? How might we generalize it to other bases?

The key is that 5 = φ(10) + 1 where φ(*n*) is the number of positive integers less than *n* and relatively prime to *n*.

Euler discovered the φ function and proved that if *a* and *m* are relatively prime, then

*a*^{φ(m)} = 1 (mod *m*)

This means that *a*^{φ(m)} − 1 is divisible by *m*. (Technically I should use the symbol ≡ (U+2261) rather than = above since the statement is a congruence, not an equation. But since Unicode font support on various devices is disappointing, not everyone could see the proper symbol.)

Euler’s theorem tells us that if *a* is relatively prime to 10 then *a*^{4} ends in 1, so *a*^{5} ends in the same digit as *a*. That proves our original statement for numbers ending in 1, 3, 7, and 9. But what about the rest, numbers that are divisible by 2 or 5? We’ll answer that question and a little more general one at the same time. Let *m* = αβ where α and β are distinct primes. In particular, we could choose α = 2 and β = 5; We will show that

*a*^{φ(m)+1} = *a* (mod *m*)

for all *a*, whether relatively prime to *m* or not. This would show in addition, for example, that in base 15, every number keeps the same last “digit” when raised to the 9th power since φ(15) = 8.

We only need to be concerned with the case of *a* being a multiple of α or β since Euler’s theorem proves our theorem for the rest of the cases. If *a* = αβ our theorem obviously holds, so assume *a* is some multiple of α, *a* = *k*α with *k* less than β (and hence relatively prime to β).

We need to show that αβ divides

(*k*α)^{φ(αβ)+1} − *k*α.

This expression is clearly divisible by α, so the remaining task is to show that it is divisible by β. We’ll show that

(*k*α)^{φ(αβ)} − 1

is divisible by β.

Since α and *k* are relatively prime to β, Euler’s theorem tells us that α^{φ(β)} and *k*^{φ(β)} are congruent to 1 (mod β). This implies that *k*α^{φ(β)} is congruent to 1, and so *k*α^{φ(α)φ(β)} is also congruent to 1 (mod β). One of the basic properties of φ is that for relatively prime arguments α and β, φ(αβ) = φ(α)φ(β) and so we’re done.

**Exercise**: How much more can you generalize this? Does the base have to be the product of two distinct primes?

**Related**: Applied number theory

Its most generalized form: The last digit of, any integer and its nth power, are the same, where n=4k+1.

This works for any squarefree base, but you can replace Euler’s totient function with Carmichael function, which gives smaller values in general (although for 10 is also 4), see https://en.wikipedia.org/wiki/Carmichael_function#Exponential_cycle_length

I first heard about this wonderful little nugget of mathematics from a YouTube video by Numberphile. It must come from the fact that the binomial expansion of the quintic power of a sum resolves beautifully as:

( a + 1 ) ^ 5 = ( a ^ 5 + 1 ) + 5 a ( a ^ 3 + 1 ) + 10 a ^ 2 ( a + 1 )

so that if a is odd or even, then ( a ^ 3 + 1 ) is even or odd, respectively,and thus { a ( a ^ 3 + 1 ) } is always even, and thus 5 times that results in a term that is a multiple of 10, which the 3rd term does by inspection, thus

( a + 1 ) ^ 5 = ( a ^ 5 + 1 ) + 10 c

[ ( a + 1 ) ^ 5 ] % 10 = ( a ^ 5 + 1 ) % 10

which means that for any given number ‘a’, that number plus 1 taken to the 5th power has the same last digit as the 5th power of the given number plus 1, which proves the theorem for any number that is +1 from some number, and thus any number can be 0 to infinity, the theorem is proven for any number from 1 to infinity. For the sum to be 0, it is trivial to prove that any power is also 0, so it works as well.

Q.E.D.

As I read through the rest of the article, I had to ask if those other equal signs are equalities or identities. It’s me, but I’m not at a point where it’s obvious.

I know when I’m entering Unicode in a figure, I don’t always get what I need, so I find some text with the symbol in it, change the font size and then screen capture it to a file of its own.