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

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(→‎Solution 1)
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Thus the answer is <math>\boxed{476}</math>.
 
Thus the answer is <math>\boxed{476}</math>.
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==Solution 2==
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The digit sum of a base <math>10</math> integer <math>m</math> is just <math>m\pmod{9}</math>. In this problem, we know <math>17\mid m</math>, or <math>m=17k</math> for a positive integer <math>k</math>.
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Also, we know that <math>m\equiv 17\equiv -1\pmod{9}</math>, or <math>17k\equiv -k\equiv -1\pmod{9}</math>.
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Obviously <math>k=1</math> is a solution. This means in general, <math>k=9x+1</math> is a solution for non-negative integer <math>x</math>.
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Checking the first few possible solutions, we find that <math>m=\boxed{476}</math> is the first solution that has <math>s(m)=17</math>, and we're done.
  
 
== 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 13:05, 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: \[\begin{array}{c|c} m & s(m)\\ \hline 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442 & 10\\ 459 & 18\\ 476 & 17 \end{array}\]

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

Solution 2

The digit sum of a base $10$ integer $m$ is just $m\pmod{9}$. In this problem, we know $17\mid m$, or $m=17k$ for a positive integer $k$.

Also, we know that $m\equiv 17\equiv -1\pmod{9}$, or $17k\equiv -k\equiv -1\pmod{9}$.

Obviously $k=1$ is a solution. This means in general, $k=9x+1$ is a solution for non-negative integer $x$.

Checking the first few possible solutions, we find that $m=\boxed{476}$ is the first solution that has $s(m)=17$, and we're done.

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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