Art of Problem Solving

2015 AIME II Problems/Problem 3: Difference between revisions

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Line 7: Line 7:
\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 $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$.

Solution 1

The three-digit integers divisible by $17$, and their digit sum: \[\begin{array}{c|c} m & s(m)\\ 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442 & 10\\ 459 & 18\\ 476 & 17 \end{array}\]

Thus the answer is $\boxed{476}$.

See also

2015 AIME II (ProblemsAnswer KeyResources)
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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