HMMT 二月 2002 · 冲刺赛 · 第 43 题

HMMT February 2002 — Guts Round — Problem 43

[9] Given that a, b, c are positive integers satisfying a + b + c = gcd( a, b ) + gcd( b, c ) + gcd( c, a ) + 120 , determine the maximum possible value of a .

解析

Given that a, b, c are positive integers satisfying a + b + c = gcd( a, b ) + gcd( b, c ) + gcd( c, a ) + 120 , determine the maximum possible value of a . Solution: 240 . Notice that ( a, b, c ) = (240 , 120 , 120) achieves a value of 240. To see that this is maximal, first suppose that a > b . Notice that a + b + c = gcd( a, b ) + gcd( b, c ) + gcd( c, a ) + 120 ≤ gcd( a, b ) + b + c + 120, or a ≤ gcd( a, b ) + 120. However, gcd( a, b ) is a proper divisor of a , so a ≥ 2 · gcd( a, b ). Thus, a − 120 ≤ gcd( a, b ) ≤ a/ 2, yielding a ≤ 240. Now, if instead a ≤ b , then either b > c and the same logic shows that b ≤ 240 ⇒ a ≤ 240, or b ≤ c, c > a (since a, b, c cannot all be equal) and then c ≤ 240 ⇒ a ≤ b ≤ c ≤ 240.