PUMaC 2014 · 数论（B 组） · 第 1 题
PUMaC 2014 — Number Theory (Division B) — Problem 1
题目详情
- [ 3 ] Let f ( x ) = x + ax + bx + c have solutions that are distinct negative integers. If a + b + c = 2014, find c. 17 17 17
解析
- [ 3 ] Let f ( x ) = x + ax + bx + c have solutions that are distinct negative integers. If a + b + c = 2014, find c. Solution: We have that x + x + x = − a , x x + x x + x x = b and x x x = − c . Thus, ( x − 1 2 3 1 2 2 3 1 3 1 2 3 1 1)( x − 1)( x − 1) = − c − b − a − 1 = − 2015. 2015 = 5 × 13 × 31, therefore the only way 2 3 that x can be distinct negative integers is when x = − 4, x = − 12, x = − 30, which gives i 1 2 3 c = 1440 . 17 17 17