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

(Created page with "==Problem== Let <math>m</math> be the least positive integer divisible by <math>17</math> whose digits sum to <math>17</math>. Find <math>m</math>. ==Solution== == See als...")
 
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Let <math>m</math> be the least positive integer divisible by <math>17</math> whose digits sum to <math>17</math>. Find <math>m</math>.
 
Let <math>m</math> be the least positive integer divisible by <math>17</math> whose digits sum to <math>17</math>. Find <math>m</math>.
  
==Solution==  
+
==Solution 1==
 +
The three-digit integers divisible by <math>17</math>, and their digit sum: <cmath>
 +
\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}
 +
</cmath>
 +
 
 +
Thus the answer is <math>\boxed{476}</math>.
  
 
== See also ==  
 
== See also ==  
 
{{AIME box|year=2015|n=II|num-b=2|num-a=4}} {{MAA Notice}}
 
{{AIME box|year=2015|n=II|num-b=2|num-a=4}} {{MAA Notice}}

Revision as of 13:00, 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:

\[\begin{array}{c|c}
m & s(m)\\
102 & 3
119 & 11
136 & 10
153 & 9
170 & 8
187 & 16
204 & 6
221 & 5
238 & 13
255 & 12
272 & 11
289 & 19
306 & 9
323 & 8
340 & 7
357 & 15
374 & 14
391 & 13
408 & 12
425 & 11
442 & 10
459 & 18
476 & 17
\end{array}\] (Error compiling LaTeX. Unknown error_msg)

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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