HMMT 二月 2009 · GEN2 赛 · 第 5 题
HMMT February 2009 — GEN2 Round — Problem 5
题目详情
- [ 4 ] Suppose a , b and c are integers such that the greatest common divisor of x + ax + b and x + bx + c is x + 1 (in the set of polynomials in x with integer coefficients), and the least common multiple of 2 2 3 2 x + ax + b and x + bx + c is x − 4 x + x + 6. Find a + b + c .
解析
- [ 4 ] Suppose a , b and c are integers such that the greatest common divisor of x + ax + b and x + bx + c is x + 1 (in the set of polynomials in x with integer coefficients), and the least common multiple of 2 2 3 2 x + ax + b and x + bx + c is x − 4 x + x + 6. Find a + b + c . Answer: − 6 2 2 Solution: Since x +1 divides x + ax + b and the constant term is b , we have x + ax + b = ( x +1)( x + b ), 2 and similarly x + bx + c = ( x + 1)( x + c ). Therefore, a = b + 1 = c + 2. Furthermore, the least common 1 3 2 multiple of the two polynomials is ( x + 1)( x + b )( x + b − 1) = x − 4 x + x + 6, so b = − 2. Thus a = − 1 and c = − 3, and a + b + c = − 6.