# 1987 AJHSME Problems/Problem 3

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## Problem $2(81+83+85+87+89+91+93+95+97+99)=$ $\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$

## Solution 1

Find that $$(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180$$ Which gives us \begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus \boxed{\text{E}} is the correct answer} \end{align*}

## Solution 2 $2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ = $83+97$ = $85+95$ = $87+93$ = $89+91$ = $180$. Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$. But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$, which is $\boxed{\text{E}}$.

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