PUMaC 2012 · 团队赛 · 第 4 题

PUMaC 2012 — Team Round — Problem 4

(3 digits) Let a = 2 + 2 and b = 2, and for n ≥ 1, a = | a − b | and b = a + b . 1 1 n +1 n n n +1 n n √ 2 2 a + a b − 6 b n n n n The minimum value of can be written in the form a b − c , where a, b, c are 2 2 6 b − a n n integers and b is square-free. Concatenate c, b, a (in that order!). 503 1006 2012

解析

Problem: (3 digits) Let a = 2 + 2 and b = 2, and for n ≥ 1, a = | a − b | and 1 1 n +1 n n √ 2 2 a + a b − 6 b n n n n b = a + b . The minimum value of can be written in the form a b − c , where n +1 n n 2 2 6 b − a n n a, b, c are integers and b is square-free. Concatenate c, b, a (in that order!). Answer: 825 Solution: First, calculate some terms in the two sequenses. Observe that ( a , b ) = (2 , 2 + 2 2 √ √ √ √ 2 2) , ( a , b ) = (2 2 , 4 + 2 2) , ( a , b ) = (4 , 4 + 4 2). Hmm... it looks like for n ≥ 2, we 3 3 4 4 √ √ n have ( a , b ) = ( 2) (1 , 1 + 2). In fact, this is true! We can prove it by induction! n n Note that 2 2 a + a b − 6 b a /b n n n n n n = − 1 + 2 2 2 6 b − a 6 − ( a /b ) n n n n so the expression only depends on the ratio a /b . We have shown that for n ≥ 2, we have: n n √ n a ( 2) 1 n = √ √ = √ n b ( 2) (1 + 2) 1 + 2 n For n = 1, we have √ √ a 2 + 2 1 √ = = 1 + 2 b 1 2 Since a /b < a /b , it follows that 2 2 1 1 a /b a /b 2 2 1 1 < 2 2 6 − ( a /b ) 6 − ( a /b ) 2 2 1 1 √ r We can see this really quickly because increases as r increases in the interval r ∈ (0 , 6). 2 6 − r Thus, the minimum value is √ a /b 2 2 − 1 + = 5 2 − 8 2 6 − ( a /b ) 2 2 Thus a = 5 , b = 2 , c = 8, and the answer is 825 Author: Alan 503 1006 2012