# A Puzzle Full of Nines

A nice puzzle by way of @benjaminleis:

This AIME problem is fun: pic.twitter.com/DhbviTqnqr

— Benjamin Leis (@benjamin_leis) February 2, 2020

In case you can’t read that, we need to find the sum of the digits in $N = 9 + 99 + 999 + 999\dots999$, where the last number consists of 321 consecutive nines.

As usual, I’ll let you pause to think about it here and post spoilers below the line.

### It feels geometric!

It feels like a geometric sequence - and it sort of is, although it took me a while to spot it. It also looks like it’s related to the decimal expansion of fractions, although that turned out to be a dead-end.

The trick (for me, at least), is to write each term as $10^k - 1$.

Then we’re looking at $N = \sum_{k=1}^{321} \left( 10^k - 1\right)$.

We can split that into two sums: a geometric sequence and $321\times1$. We *could* bring out the formula for that, but why would we? It’s going to be a string of 321 ones with a zero at the end.

From that, we need to subtract 321 - and we only really care about the last four digits here: $1110 - 321 = 0789$.

So, we’ve still got 318 ones in our string (the digits of which sum to 318), followed by 0789 (the digits of which sum to 24), making a total of 342.

### Quick check

The number $N$ is a multiple of 9, so its digits sum to a multiple of 9. It’s worth checking that our answer makes sense: is 342 a multiple of 9? Well, its digits sum to 9, so it looks plausible! (It’s $38\times 9$, the Ninja murmurs in their sleep).

It doesn’t mean we definitely have the right answer - but it increases our confidence slightly; had we got something that wasn’t a multiple of 9, we’d know we had it wrong.

I enjoyed that puzzle. Did you tackle it a different way?