Difference between revisions of "2015 AIME II Problems/Problem 3"
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\begin{array}{c|c} | \begin{array}{c|c} | ||
m & s(m)\\ | m & s(m)\\ | ||
| − | 102 & 3 | + | 102 & 3 \\ |
| − | 119 & 11 | + | 119 & 11\\ |
| − | 136 & 10 | + | 136 & 10\\ |
| − | 153 & 9 | + | 153 & 9\\ |
| − | 170 & 8 | + | 170 & 8\\ |
| − | 187 & 16 | + | 187 & 16\\ |
| − | 204 & 6 | + | 204 & 6\\ |
| − | 221 & 5 | + | 221 & 5\\ |
| − | 238 & 13 | + | 238 & 13\\ |
| − | 255 & 12 | + | 255 & 12\\ |
| − | 272 & 11 | + | 272 & 11\\ |
| − | 289 & 19 | + | 289 & 19\\ |
| − | 306 & 9 | + | 306 & 9\\ |
| − | 323 & 8 | + | 323 & 8\\ |
| − | 340 & 7 | + | 340 & 7\\ |
| − | 357 & 15 | + | 357 & 15\\ |
| − | 374 & 14 | + | 374 & 14\\ |
| − | 391 & 13 | + | 391 & 13\\ |
| − | 408 & 12 | + | 408 & 12\\ |
| − | 425 & 11 | + | 425 & 11\\ |
| − | 442 & 10 | + | 442 & 10\\ |
| − | 459 & 18 | + | 459 & 18\\ |
476 & 17 | 476 & 17 | ||
\end{array} | \end{array} | ||
Revision as of 13:01, 26 March 2015
Problem
Let 
be the least positive integer divisible by 
whose digits sum to . Find .
Solution 1
The three-digit integers divisible by , and their digit sum: ![\[\begin{array}{c|c} m & s(m)\\ 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442 & 10\\ 459 & 18\\ 476 & 17 \end{array}\]](http://latex.artofproblemsolving.com/1/1/7/1174e4a113dade1511a28297193c95e4718a456d.png)
Thus the answer is 
.
See also
| 2015 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
