2015 AIME II Problems/Problem 3

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.

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

Page

Toolbox

Search

2015 AIME II Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also