Difference between revisions of "1992 AJHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | The | + | The highest digit sum for three-digit numbers is <math> 9+9+9=27 </math>. Therefore, the only possible digit combination is <math> 9, 9, 8 </math>. Of course, of the three possible numbers, only <math> 998 </math> works. Thus, the answer is <math> \boxed{\text{(A)}\ 1} </math>. |
==See Also== | ==See Also== | ||
| + | {{AJHSME box|year=1992|num-b=6|num-a=8}} | ||
| − | + | [[Category:Introductory Algebra Problems]] | |
| − | [[Category:Introductory | + | {{MAA Notice}} |
Latest revision as of 14:45, 24 May 2017
Problem
The digit-sum of 
is 
. How many 3-digit whole numbers, whose digit-sum is 
, are even?
Solution
The highest digit sum for three-digit numbers is 
. Therefore, the only possible digit combination is 
. Of course, of the three possible numbers, only works. Thus, the answer is 
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
