2016 AIME I Problems/Problem 8

Problem 8

For a permutation $p = (a_1,a_2,\ldots,a_9)$ of the digits $1,2,\ldots,9$ , let $s(p)$ denote the sum of the three $3$ -digit numbers $a_1a_2a_3$ , $a_4a_5a_6$ , and $a_7a_8a_9$ . Let $m$ be the minimum value of $s(p)$ subject to the condition that the units digit of $s(p)$ is $0$ . Let $n$ denote the number of permutations $p$ with $s(p) = m$ . Find $|m - n|$ .

Solution

Solution by jonnyboyg: To minimize $s(p)$ , the numbers 1, 2, and 3 must be in the hundreds places. The numbers in the ones places must have a sum of 20. This means that the numbers in the tens places will always have the same sum. One way to do this is 154+267+389=810. Therefore m=810.

To find n, realize that there are 3!=6 ways of ordering the numbers in each of the places. Additionally, there are three possibilities for the numbers in the ones place-4 7 and 9, 5 7 and 8, and 6 7 and 9. Therefore there are 6^3 * 3 ways total, which is 648. $|m-n|$ = $|810-648|$ =162.

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2016 AIME I Problems/Problem 8

Problem 8

Solution