PUMaC 2015 · 代数（B 组） · 第 4 题

PUMaC 2015 — Algebra (Division B) — Problem 4

[ 4 ] There are real numbers a, b, c, d such that for all ( x, y ) satisfying 6 y = 2 x + 3 x + x , if 2 3 x = ax + b and y = cy + d , then y = x − 36 x . What is a + b + c + d ? 1 1 1 1 1 6 5 4 3

解析

[ 4 ] There are real numbers a, b, c, d such that for all ( x, y ) satisfying 6 y = 2 x + 3 x + x , if 2 3 x = ax + b and y = cy + d , then y = x − 36 x . What is a + b + c + d ? 1 1 1 1 1 Solution: 2 3 2 ′ ′ We have 6 y = 2 x + 3 x + x . Make the following substitution: y = 3 y , x = 3 x . This gives 1 1 1 ′ 2 ′ 3 ′ 2 ′ ′′ ′ y = x + x + x . Further substitute x = x + . This gives 2 18 6 1 ′ 2 ′′ 3 ′′ y = x − x 36 3 ′ 2 3 ′′ 3 2 ′′ 36 y = 36 x − 36 x 3 ′ 2 ′′ 3 ′′ (6 y ) = (36 x ) − 36(36 x ) . ′ ′′ 2 3 Finally, make the substitution y = 216 y and x = 36 x . Thus y = x − 36 x , and by 1 1 1 1 1 substitution, y = 72 y and x = 12 x + 6. So our answer is 12 + 6 + 72 + 0 = 90 . 1 1 EDIT: This problem did not specify that a and c should be non-zero. As such, any solution 2 3 using a = c = 0 and ( b, d ) a solution to y = x − 36 x was valid. This problem has been 1 1 1 thrown out due to such a wide variety of possible answers. Author: Heesu Hwang 6 5 4 3