# Difference between revisions of "1985 AJHSME Problems/Problem 2"

## Problem

$90+91+92+93+94+95+96+97+98+99=$

$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

## Solution

We could just add them all together. But what would the point be of doing that? So we find a slicker way.

We find a simpler problem in this problem, and simplify -> $90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9$
We know $90 \times 10$, that's easy - $900$. So how do we find $1 + 2 + ... + 8 + 9$?

Well once again (not once again if you didn't read the solution for #1), commutative property comes to the rescue. We rearrange the numbers to make $(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5$. You might have noticed that each of the terms i put next to each other add up to 10, which make for easy adding. $4 \times 10 + 5 = 45$. Adding that on to 900 makes 945.

945 is (B)