Find ones digit of $1^1+2^2+3^3+\cdots+117^{117}$

Asked Feb 27 '22 at 10:07

Active Feb 27 '22 at 10:07

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Find the ones digit of

$$1^1+2^2+3^3+\cdots+117^{117}$$

My work:

The problem is equivalent as finding $1^1+2^2+3^3+\cdots+117^{117}$ modulo 10.

I tried Euler's theorem, finding $$a^4\equiv 1 \text{ (mod 10)}$$

However, the problem is that it only works for $a$ that is coprime with $10$. So we can't find a pattern easily this way.

Please help!

asked Feb 27 '22 at 10:07

Cyh1368

You can similarly show e.g. $2^{n+4}\equiv2^n$ for $n\ge1$. – J.G. Feb 27 '22 at 10:17
6

It appears that the sequence $n^n\pmod{10}$ is periodic with period $20$. – pancini Feb 27 '22 at 10:21
Hint: $(1), (2,4,8,6),(3,9,7,1),(4,6),(5),(6),(7,9,3,1),(8,4,2,6),(9,1),(0)$ – jjagmath Feb 27 '22 at 11:41
1

$a^5\equiv a \bmod 10$ for all $a$, even when $\gcd(a,10)>1$. – lhf Feb 27 '22 at 12:08
@lhf I suppose that's from Fermat's Little Theorem, but it should only hold if $10$ is a prime, which it isn't. – Cyh1368 Feb 27 '22 at 15:20
No, it's a general result: https://en.wikipedia.org/wiki/Carmichael_function#Exponential_cycle_length – lhf Feb 27 '22 at 15:30

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