Question

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 $|m - n|.$

Answer

Solution:

By place value, $s(p) = 100(a_1 + a_4 + a_7) + 10(a_2 + a_5 + a_8) + (a_3 + a_6 + a_9),$ and all nine digits sum to $45.$ The units digit of $s(p)$ is $0$ exactly when $a_3 + a_6 + a_9 = 10$ or $20.$ Writing $X = a_1 + a_4 + a_7,$ if the units column sums to $10$ then $s(p) = 100X + 10(35 - X) + 10 = 90X + 360 \ge 900,$ while if it sums to $20$ then $s(p) = 90X + 270 \ge 90 \cdot 6 + 270 = 810.$ So $m = 810,$ achieved exactly when $\{a_1, a_4, a_7\} = \{1, 2, 3\}$ and the units digits sum to $20.$

The remaining digits $\{4, 5, 6, 7, 8, 9\}$ must split so the units triple sums to $20:$ the possibilities are $\{4, 7, 9\},$ $\{5, 6, 9\},$ and $\{5, 7, 8\}.$ Each of the $3$ splits allows $3! \cdot 3! \cdot 3! = 216$ arrangements of the three columns, so $n = 3 \cdot 216 = 648.$

Therefore $|m - n| = |810 - 648| = 162.$

2016 AIME I Problem 8

Solution:

Problem 8 in Other Years