Difference between revisions of "2015 AIME II Problems/Problem 3"

(Solution)
(Solution 1)
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 14:01, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png