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 $\boxed{\text{(C)}\ 6}$ .

1994 AJHSME (Problems • Answer Key • Resources)
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.

Page

Toolbox

Search

1994 AJHSME Problems/Problem 8

Problem

Solution

See Also