Problem:
Let a,b,c,d,e,f,g,h,i be distinct integers from 1 to 9 . The minimum possible positive value of
g⋅h⋅ia⋅b⋅c−d⋅e⋅f
can be written as nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Answer (289):
First consider the case when abc=def+1. Let X=abc. Then
ghiabc−def=ghi1=9!abcdef=9!X⋅(X−1).
Because X>1, this is an increasing function of X. Note that X(X−1)=abcdef≥6!, and therefore X≥28. Either X or X−1 is odd and is therefore a product of three distinct odd factors from {1,2,…,9}.\
If X−1=27, then {d,e,f}={1,3,9}. However, X=28 is not factorable into three distinct positive integers each greater than 1 . The next odd number greater than 27 whose prime factors are less than 10 is 35=7⋅5⋅1. If either X or X−1 is 35 , then the other is either 34 or 36 . But 17 is a prime factor of 34 , so this forces X to be 36=6⋅3⋅2.\
Therefore in the case when abc and def are consecutive integers, the minimum value of the expression is
9!36⋅35=9⋅8⋅46⋅3⋅2−7⋅5⋅1=2881
If abc≥def+2, then
ghiabc−def≥9⋅8⋅72=2521>2881
Hence the least possible positive value of the expression is 2881. The requested sum is 1+288=289.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions