1990 AJHSME Problems/Problem 16

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Problem

$1990-1980+1970-1960+\cdots -20+10 =$

$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$

Solution

In the middle, we have $\cdots + 1010-1000+990 -\cdots$ .

If we match up the back with the front, and then do the same for the rest, we get pairs with $2000$ and $-2000$ , so these will cancel out. In the middle, we have $2000-1000$ which doesn't cancel, but gives us $1000 \rightarrow \boxed{\text{D}}$ .

1990 AJHSME (Problems • Resources)
Preceded by Problem 15	Followed by Problem 17
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1990 AJHSME Problems/Problem 16

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Problem

Solution

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