# Puzzle with sums and powers

If the sum of the digits of 77^77 is S, and the sum of the digits of S is T, find the sum of the digits of T.

## Computational solution

Python says (since wasn't applied):

$$77^{77}=$$18188037387806198379277339915556929647807403283187048631478337739929618787870634227045716719924575689062274471430368865388203540672666042530996797
$$S=722$$
$$T=11$$
$$\text{Sum of digits of T }=2$$

## Manual solution

First:

$$77^{77}<100^{100}$$, so the digit sum ($$S$$) is at most $$1800$$. But then the digit sum of that ($$T$$) is at most $$27$$. Let $$U$$ be the digit sum of $$T$$, then $$U$$ is at most $$10$$.

Also:

$$77\equiv5\text{ (mod }9\text)$$. In $$\text{mod }9$$, the powers of $$5$$ go $$5,7,8,4,2,1,5,7,8,4,2,1,\dots$$ and so since this cycle is of length $$6$$ and $$77\equiv5\text{ (mod }6\text)$$, $$77^{77}\equiv2\text{ (mod }9\text)$$ (the second last term in the cycle). It is well known that the digit sum is equivalent to the original sum $$\text{(mod }9\text)$$, so $$U\equiv T\equiv S\equiv2\text{ (mod }9\text)$$ as well.

So $$U$$ is:

$$2$$, since that is the only number $$\leq10$$ that is $$\equiv2\text{ (mod }9\text)$$

• The question asked for the sum of digits of T, not T.
– ffao
Sep 21, 2017 at 1:41
• @ffao I should lose 100 reputation each time I don't read the question carefully, maybe that would teach me to read things through. Sep 21, 2017 at 2:40