Problem:
For a permutation p=(a1,a2,…,a9) of the digits 1,2,…,9, let s(p) denote the sum of the three 3-digit numbers a1a2a3,a4a5a6, and a7a8a9. 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:
For any permutation p
s(p)=100(a1+a4+a7)+10(a2+a5+a8)+(a3+a6+a9)
Because the units digit of m is 0, it follows that a3+a6+a9=10 or 20.
If a3+a6+a9=10, then a1+a4+a7+a2+a5+a8=45−10=35, and s(p)=90(a1+a4+a7)+360≥90(1+2+3)+360=900.
If a3+a6+a9=20, then s(p)=90(a1+a4+a7)+270≥90⋅6+270=810, with equality if and only if {a1,a4,a7}={1,2,3}. The set {a3,a6,a9} must equal {4,7,9},{5,6,9}, or {5,7,8}.
For each {ai,ai+3,ai+6}, there are 3! ways to permute these numbers. Hence there are n=3⋅3!⋅3!⋅3!=648 permutations p with s(p)=m=810 and ∣m−n∣=∣810−648∣=162.
The problems on this page are the property of the MAA's American Mathematics Competitions