1994 AJHSME Problems/Problem 8

Problem

For how many three-digit whole numbers does the sum of the digits equal $25$?

$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Solution

Because $8+8+8=24$, it follows that one of the digits must be a $9$. The other two digits them have a sum of $25-9=16$. The groups of digits that produce a sum of $25$ are $799, 889$ and can be arranged as follows

\[799,979,997,889,898,988\]

The number of configurations is $\boxed{\text{(C)}\ 6}$.

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS