2016 AIME I 第 8 题

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8.

对数字 1,2,,91, 2, \ldots, 9 的一个排列 p=(a1,a2,,a9)p = (a_1, a_2, \ldots, a_9),令 s(p)s(p) 表示三个 33 位数 a1a2a3a_1a_2a_3a4a5a6a_4a_5a_6a7a8a9a_7a_8a_9 的和。设 mm 为所有满足 s(p)s(p) 的个位数字为 00 的排列中,s(p)s(p) 的最小值。设 nn 表示满足 s(p)=ms(p) = m 的排列 pp 的个数。求 mn|m - n|

For a permutation p=(a1,a2,,a9)p = (a_1, a_2, \ldots, a_9) of the digits 1,2,,9,1, 2, \ldots, 9, let s(p)s(p) denote the sum of the three 33-digit numbers a1a2a3,a_1a_2a_3, a4a5a6,a_4a_5a_6, and a7a8a9.a_7a_8a_9. Let mm be the minimum value of s(p)s(p) subject to the condition that the units digit of s(p)s(p) is 0.0. Let nn denote the number of permutations pp with s(p)=m.s(p) = m. Find mn.|m - n|.

答案:162
知识点:位值排列分类讨论
难度评级:2710
解答:

按位值,s(p)=100(a1+a4+a7)s(p) = 100(a_1 + a_4 + a_7) +10(a2+a5+a8)+ 10(a_2 + a_5 + a_8) +(a3+a6+a9)+ (a_3 + a_6 + a_9),且九个数字之和为 4545s(p)s(p) 的个位数字为 00 当且仅当 a3+a6+a9=10a_3 + a_6 + a_9 = 102020。令 X=a1+a4+a7X = a_1 + a_4 + a_7,如果个位列和为 1010,则 s(p)=100X+10(35X)+10s(p) = 100X + 10(35 - X) + 10 =90X+360900= 90X + 360 \ge 900;如果个位列和为 2020,则 s(p)=90X+270s(p) = 90X + 270 906+270=810\ge 90 \cdot 6 + 270 = 810。所以 m=810m = 810,恰好在 {a1,a4,a7}={1,2,3}\{a_1, a_4, a_7\} = \{1, 2, 3\} 且个位数字之和为 2020 时达到。

剩余数字 {4,5,6,7,8,9}\{4, 5, 6, 7, 8, 9\} 必须拆分,使个位三元组和为 2020:可能为 {4,7,9}\{4, 7, 9\}{5,6,9}\{5, 6, 9\}, 和 {5,7,8}\{5, 7, 8\}。这 33 种拆分中的每一种都允许 3!3!3!=2163! \cdot 3! \cdot 3! = 216 种三列排列,所以 n=3216=648n = 3 \cdot 216 = 648

因此 mn=810648=162|m - n| = |810 - 648| = 162

By place value, s(p)=100(a1+a4+a7)s(p) = 100(a_1 + a_4 + a_7) +10(a2+a5+a8)+ 10(a_2 + a_5 + a_8) +(a3+a6+a9),+ (a_3 + a_6 + a_9), and all nine digits sum to 45.45. The units digit of s(p)s(p) is 00 exactly when a3+a6+a9=10a_3 + a_6 + a_9 = 10 or 20.20. Writing X=a1+a4+a7,X = a_1 + a_4 + a_7, if the units column sums to 1010 then s(p)=100X+10(35X)+10s(p) = 100X + 10(35 - X) + 10 =90X+360900,= 90X + 360 \ge 900, while if it sums to 2020 then s(p)=90X+270s(p) = 90X + 270 906+270=810.\ge 90 \cdot 6 + 270 = 810. So m=810,m = 810, achieved exactly when {a1,a4,a7}={1,2,3}\{a_1, a_4, a_7\} = \{1, 2, 3\} and the units digits sum to 20.20.

The remaining digits {4,5,6,7,8,9}\{4, 5, 6, 7, 8, 9\} must split so the units triple sums to 20:20: the possibilities are {4,7,9},\{4, 7, 9\}, {5,6,9},\{5, 6, 9\}, and {5,7,8}.\{5, 7, 8\}. Each of the 33 splits allows 3!3!3!=2163! \cdot 3! \cdot 3! = 216 arrangements of the three columns, so n=3216=648.n = 3 \cdot 216 = 648.

Therefore mn=810648=162.|m - n| = |810 - 648| = 162.

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