HMMT 二月 2015 · 冲刺赛 · 第 15 题
HMMT February 2015 — Guts Round — Problem 15
题目详情
- [ 8 ] Find the maximum possible value of H · M · M · T over all ordered triples ( H, M, T ) of integers such that H · M · M · T = H + M + M + T .
解析
- [ 8 ] Find the maximum possible value of H · M · M · T over all ordered triples ( H, M, T ) of integers such that H · M · M · T = H + M + M + T . Answer: 8 If any of H, M, T are zero, the product is 0. We can do better (examples below), so we may now restrict attention to the case when H, M, T 6 = 0. When M ∈ {− 2 , − 1 , 1 , 2 } , a little casework gives all the possible ( H, M, T ) = (2 , 1 , 4) , (4 , 1 , 2) , ( − 1 , − 2 , 1) , (1 , − 2 , − 1). • If M = − 2, i.e. H − 4 + T = 4 HT , then − 15 = (4 H − 1)(4 T − 1), so 4 H − 1 ∈ {± 1 , ± 3 , ± 5 , ± 15 } (only − 1 , +3 , − 5 , +15 are possible) corresponding to 4 T − 1 ∈ {∓ 15 , ∓ 5 , ∓ 3 , ∓ 1 } (only +15 , − 5 , +3 , − 1 are possible). But H, T are nonzero, we can only have 4 H − 1 ∈ { +3 , − 5 } , yielding ( − 1 , − 2 , 1) and (1 , − 2 , − 1). • If M = +2, i.e. H + 4 + T = 4 HT , then 17 = (4 H − 1)(4 T − 1), so 4 H − 1 ∈ {± 1 , ± 17 } (only − 1 , − 17 are possible) corresponding to 4 T − 1 ∈ {± 17 , ± 1 } (only − 17 , − 1 are possible). But H, T are nonzero, so there are no possibilities here. • If M = − 1, i.e. H − 2 + T = HT , then − 1 = ( H − 1)( T − 1), so we have H − 1 ∈ {± 1 } and T − 1 ∈ {∓ 1 } , neither of which is possible (as H, T 6 = 0). • If M = +1, i.e. H + 2 + T = HT , then 3 = ( H − 1)( T − 1), so we have H − 1 ∈ {± 1 , ± 3 } . Since H, T 6 = 0, H − 1 ∈ { +1 , +3 } , yielding (2 , 1 , 4) and (4 , 1 , 2). 2 Now suppose there is such a triple ( H, M, T ) for | M | ≥ 3. The equation in the problem gives ( M H − 2 3 3 2 2 2 1)( M T − 1) = 2 M + 1. Note that since H, T 6 = 0, | 2 M + 1 | = | M H − 1 | · | M T − 1 | ≥ min( M − 2 2 4 2 3 1 , M + 1) = M − 2 M + 1 > 2 | M | + 1 gives a contradiction.