# 2019 AIME I Problems/Problem 1

## Problem 1

Consider the integer $$N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.$$Find the sum of the digits of $N$.

## Solution 1

Let's express the number in terms of $10^n$. We can obtain $(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)$. By the commutative and associative property, we can group it into $(10+10^2+10^3+\cdots+10^{321})-321$. We know the former will yield $1111....10$, so we only have to figure out what the last few digits are. There are currently $321$ 1's. We know the last four digits are $1110$, and that the others will not be affected if we subtract $321$. If we do so, we get that $1110-321=789$. This method will remove three $1$'s, and add a $7$, $8$ and $9$. Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$.

-eric2020

A similar and simpler way to consider the initial manipulations is to observe that adding $1$ to each term results in $(10+100+... 10^{320}+10^{321})$. There are $321$ terms, so it becomes $11...0$, where there are $322$ digits in $11...0$. Then, subtract the $321$ you initially added.

~ BJHHar

## Solution 2

We can see that $9=9$, $9+99=108$, $9+99+999=1107$, all the way to ten nines when we have $11111111100$. Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is $\boxed{342}$ since we have to add $9\lfloor \log 321 \rfloor$ to the sum of digits, which is $9\lceil \frac{321}9 \rceil$.

## Solution 3 (Pattern)

Observe how adding results in the last term but with a $1$ concatenated in front and also a $1$ subtracted ( $09$, $108$, $1107$, $11106$). Then for any index of terms, $n$, the sum is $11...10-n$, where the first term is of length $n+1$. Here, that is $\boxed{342}$.

~BJHHar

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 