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Difference between revisions of "1992 AJHSME Problems/Problem 7"

(Created page with "== Problem 7 == The digit-sum of <math>998</math> is <math>9+9+8=26</math>. How many 3-digit whole numbers, whose digit-sum is <math>26</math>, are even? <math>\text{(A)}\ 1 \...")
 
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== Problem 7 ==
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== Problem==
  
 
The digit-sum of <math>998</math> is <math>9+9+8=26</math>.  How many 3-digit whole numbers, whose digit-sum is <math>26</math>, are even?
 
The digit-sum of <math>998</math> is <math>9+9+8=26</math>.  How many 3-digit whole numbers, whose digit-sum is <math>26</math>, are even?

Revision as of 15:24, 29 July 2011

Problem

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 $\boxed{\text{(A)}\ 1}$.

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