Difference between revisions of "2015 AIME II Problems/Problem 3"
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The three-digit integers divisible by <math>17</math>, and their digit sum: <cmath> | The three-digit integers divisible by <math>17</math>, and their digit sum: <cmath> | ||
\begin{array}{c|c} | \begin{array}{c|c} | ||
| − | m & s(m)\\ | + | m & s(m)\\ \hline |
102 & 3 \\ | 102 & 3 \\ | ||
119 & 11\\ | 119 & 11\\ | ||
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: ![\[\begin{array}{c|c} m & s(m)\\ \hline 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442 & 10\\ 459 & 18\\ 476 & 17 \end{array}\]](http://latex.artofproblemsolving.com/1/f/e/1feb56dcf7d45963aa10db2a20aeed303e13f7a5.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. 
