HMMT 二月 2024 · 冲刺赛 · 第 20 题

HMMT February 2024 — Guts Round — Problem 20

[11] Compute 5508 + 5625 + 5742 , given that it is an integer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2024, February 17, 2024 — GUTS ROUND Organization Team Team ID#

解析

[11] Compute 5508 + 5625 + 5742 , given that it is an integer. Proposed by: Rishabh Das Answer: 855 2 Solution: Let a = 5625 = 75 and b = 117 . Then we have 3 3 3 3 3 3 3 2 2 2 5508 + 5265 + 5742 = ( a − b ) + a + ( a + b ) = 3 a + 6 ab = 3 a ( a + 2 b ) . 3 4 2 2 4 2 2 We have 3 a = 3 · 5 , so a + 2 b = 3 · (625 + 2 · 19 ) should be 3 times a fourth power. This means 2 2 4 625 + 2 · 19 = 3 x √ √ 2 for some integer x . By parity, x must be odd, and also x 3 ≈ 625 . Approximating 3 even as 2 , we get x should be around 19 . Then x = 17 is clearly too small, and x = 21 is too big. (You can also check mod 7 for this latter one.) Thus, x = 19 . The final answer is then 2 3 · 5 · 19 = 855 .