Difference between revisions of "1994 AJHSME Problems/Problem 8"
Mrdavid445 (talk | contribs) (Created page with "==Problem== For how many three-digit whole numbers does the sum of the digits equal <math>25</math>? <math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{...") |
|||
| Line 4: | Line 4: | ||
<math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math> | <math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math> | ||
| + | |||
| + | ==Solution== | ||
| + | Because <math>8+8+8=24</math>, it follows that one of the digits must be a <math>9</math>. The other two digits them have a sum of <math>25-9=16</math>. The groups of digits that produce a sum of <math>25</math> are <math>799, 889</math> and can be arranged as follows | ||
| + | |||
| + | <cmath>799,979,997,889,898,988</cmath> | ||
| + | |||
| + | The number of configurations is <math>\boxed{\text{(C)}\ 6}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=7|num-a=9}} | ||
Revision as of 23:24, 22 December 2012
Problem
For how many three-digit whole numbers does the sum of the digits equal 
?
Solution
Because 
, it follows that one of the digits must be a 
. The other two digits them have a sum of 
. The groups of digits that produce a sum of are 
and can be arranged as follows
![\[799,979,997,889,898,988\]](http://latex.artofproblemsolving.com/6/1/d/61d91bf433513a2be7ef32a4fcb9b07005778dff.png)
The number of configurations is 
.
See Also
| 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 | ||
