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 (Faster)

We can do the same thing as solution 1, except note the following fact: $102$ is a multiple of $17$ and its digits sum to $3$ .

Therefore, we can add it onto an existing multiple of $17$ that we know of to have $s(m) = 14$ , shown in the right-hand column, provided that its units digit is less than $8$ and its hundreds digit is less than $9$ . Unfortunately, $68$ does not fit the criteria, but $374$ does, meaning that, instead of continually adding multiples of $17$ , we can stop here and simply add $102$ to reach our final answer of $\boxed{476}$ .

~Tiblis

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.

Solution 3

Since the sum of the digits in the base-10 representation of $m$ is $17$ , we must have $m\equiv 17 \pmod{9}$ or $m\equiv -1\pmod{9}$ . We also know that since $m$ is divisible by 17, $m\equiv 0 \pmod{17}$ .

To solve this system of linear congruences, we can use the Chinese Remainder Theorem. If we set $m\equiv (-1)(17)(8)\pmod {153}$ , we find that $m\equiv 0\pmod{17}$ and $m\equiv -1\pmod{9}$ , because $17\cdot 8\equiv 136 \equiv 1\pmod{9}$ . The trick to getting here was to find the number $x$ such that $17x\equiv 1\pmod{9}$ , so that when we take things $\pmod{9}$ , the $17$ goes away. We can do this using the Extended Euclidean Algorithm or by guess and check to find that $x\equiv 8\pmod{9}$ .

Finally, since $m\equiv 17\pmod{153}$ , we repeatedly add multiples of $153$ until we get a number in which its digits sum to 17, which first happens when $m=\boxed{476}$ .

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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