HMMT 十一月 2023 · 冲刺赛 · 第 7 题

7. [7] Suppose a and b be positive integers not exceeding 100 such that 2 lcm( a, b ) ab = . gcd( a, b ) Compute the largest possible value of a + b . Proposed by: Rishabh Das Answer: 78 Solution: For any prime p and a positive integer n , let ν ( n ) be the largest nonnegative integer k for p k which p divides n . Taking ν on both sides of the given equation, we get p ν ( a ) + ν ( b ) = 2 · | ν ( a ) − ν ( b ) | , p p p p ν ( a ) p 1 which means ∈ 3 , for all primes p . Using this with a, b ≤ 100, we get that ν ( b ) 3 p • We must have ( ν ( a ) , ν ( b )) ∈ { (0 , 0) , (1 , 3) , (3 , 1) , (2 , 6) , (6 , 2) } because a and b cannot be divisible 2 2 7 by 2 . 6 • We must have ( ν ( a ) , ν ( b )) ∈ { (0 , 0) , (1 , 3) , (3 , 1) } because a and b cannot be divisible by 3 > 100. 3 3 • a and b cannot be divisible by any prime p ≥ 5, because if not, then one of a and b must be 3 3 divisible by p ≥ 5 > 100. If ( ν ( a ) , ν ( b )) = (2 , 6) (and similarly with (6 , 2)), then we must have ( a, b ) = (4 , 64), so the sum is 2 2

If ( ν ( a ) , ν ( b )) = (1 , 3) (and similarly with (3 , 1)), then we must have ν ( b ) ≤ 1 (otherwise, b ≥ 3 3 2 2 3 3 1 1 3 2 · 3 > 100). Hence, the optimal pair is ( a, b ) = (2 · 3 , 2 · 3 ) = (24 , 54), so the sum is 24 + 54 = 78. 1 3 If neither of the above happens, then a + b ≤ 2 + 2 ≤ 10, which is clearly not optimal. Hence, the optimal pair is (24 , 54), and the answer is 78.