1990 AJHSME Problems/Problem 16

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 $1000 \rightarrow \boxed{\text{D}}$.

Solution 2

We can see that there are $199$ terms in total. We can also see that the first $198$ numbers form groups of two that add to $10$ each. Dividing to see how many pairs we have, $198$/$2$ = $99$ groups of ten, or $990$. However, we have to remember to add the 199th term ($10$), so we get $990$ + $10$ = $1000$, which gives us $\boxed{\text{D}}$.

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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