2015 AIME II Problems/Problem 3: Difference between revisions
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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:
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 | ||
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