1992 AJHSME Problems/Problem 7

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Problem 7

The digit-sum of $998$ is $9+9+8=26$. How many 3-digit whole numbers, whose digit-sum is $26$, are even?

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution

The hightest digit sum for three-digit numbers is $9+9+9=27$. Therefore, the only possible digit combination is $9, 9, 8$. Of course, of the three possible numbers, only $998$ works. Thus, the answer is $\boxed{\text{(A)}\ 1}$.