HMMT 十一月 2020 · 冲刺赛 · 第 10 题

HMMT November 2020 — Guts Round — Problem 10

[8] The number 3003 is the only number known to appear eight times in Pascal’s triangle, at positions ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3003 3003 a a 15 15 14 14 , , , , , , , . 1 3002 2 a − 2 b 15 − b 6 8 Compute a + b (15 − b ).

解析

[8] The number 3003 is the only number known to appear eight times in Pascal’s triangle, at positions ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3003 3003 a a 15 15 14 14 , , , , , , , . 1 3002 2 a − 2 b 15 − b 6 8 Compute a + b (15 − b ). Proposed by: Carl Joshua Quines Answer: 128 ( ) 2 a a ( a − 1) a Solution: We first solve for a . Note that 3003 = 3 · 7 · 11 · 13. We have 3003 = = ≈ . This 2 2 2 √ means we can estimate a ≈ 3003 · 2, so a is a little less than 80. Furthermore, 11 | 2 · 3003 = a ( a − 1), meaning one of a or a − 1 must be divisible by 11. Thus, either a = 77 or a = 78. Conveniently, 13 | 78, ( ) 78 so we get a = 78 and we can verify that = 3003. 2 ( ) ( ) 15 15 15! We solve for b < 15 − b satisfying = 3003. Because = is divisible by 11, we must b b b !(15 − b )! ( ) 14 have b ≥ 5. But we’re given = 3003, so b < 6. We conclude that b = 5, and it follows that 6 a + b (15 − b ) = 128.