PUMaC 2015 · 团队赛 · 第 10 题

PUMaC 2015 — Team Round — Problem 10

[ 7 ] Let S be the set of integer triplets ( a, b, c ) with 1 ≤ a ≤ b ≤ c that satisfy a + b + c = 77 and: 1 1 1 1

解析

[ 7 ] Let S be the set of integer triplets ( a, b, c ) with 1 ≤ a ≤ b ≤ c that satisfy a + b + c = 77 and: 1 1 1 1

- = . a b c 5 ∑ What is the value of the sum a · b · c ? ( a,b,c ) ∈ S Solution: We multiply both sides of the sum of fractions equation by 5 abc to get: 5( ab + bc + ac ) = abc Now adding 25( a + b + c ) − 125 to both sides, we get that: 2 3 2 2 ( a − 5)( b − 5)( c − 5) = 25(77) − 125 = 5 · 72 = 2 · 3 · 5 3 2 2 Letting x = a − 5 , y = b − 5 , z = c − 5, we want x + y + z = 62 and xyz = 2 · 3 · 5 . First, by the fraction equation, we know that a > 5 and so 1 ≤ x, y, z . Then either 5 divides two of x, y, z or 25 divides one of x, y, z . Case 1 : 5 divides two of x, y, z : Let w denote the number out of the triplet x, y, z not divisible by 5. Since xyz = 72 · 25, w must divide 72. But since x + y + z = 62 ≡ 2 mod 5 and 5 divides the other two integers, w ≡ 2 mod 5. So we can see that w = 2 , 12 , or 72. But x + y + z = 62 so w 6 = 72. Case 1 a : w = 2 : If w = 2, then the other two numbers are of the form 5 · m, 5 · n where 60 m + n = = 12 and mn = 36. So m = n = 6 and ( x, y, z ) = (2 , 30 , 30) solves the two 5 equation. Case 1 b : w = 12 : If w = 2, then the other two numbers are of the form 5 · m, 5 · n where 50 m + n = = 10 and mn = 6. So there are no solutions with w = 12. 5 Case 2 : 25 divides one of x, y, z : Let w denote the number of the triplet that is divisible by