AMC12 2004 B

AMC12 2004 B · Q23

AMC12 2004 B · Q23. It mainly tests Vieta / quadratic relationships (basic), Number theory misc.

The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?

多项式 $x^3 - 2004 x^2 + mx + n$ 的系数为整数，且有三个互不相同的正零点。其中恰好有一个零点是整数，并且它等于另外两个零点之和。问 $n$ 可能有多少个取值？

(A) 250000 250000

(B) 250250 250250

(D) 250750 250750

(E) 251000 251000

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let the roots be $r,s,r + s$, and let $t = rs$. Then $(x - r)(x - s)(x - (r + s))$ $= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0$ and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$. Then our polynomial looks like \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] and we need the number of possible products $t = rs = r(1002 - r)$. Because $m=t+1002^2$ is an integer, we also note that $t$ must be an integer. Since $r > 0$ and $t > 0$, it follows that $0 < t = r(1002-r) < 501^2 = 251001$, with the endpoints not achievable because the roots must be distinct and positive. Because neither $r$ nor $1002-r$ can be an integer, there are $251,000 - 500 = \boxed{\textbf{(C) } 250,500}$ possible values of $n = -1002t$.

设三个根为 $r,s,r+s$，并令 $t=rs$。则 $(x - r)(x - s)(x - (r + s))$ $= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0$ 对比系数得 $2(r+s)=2004 \Longrightarrow r+s=1002$。因此多项式为 \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] 我们需要计算可能的乘积 $t=rs=r(1002-r)$ 的个数。由于 $m=t+1002^2$ 为整数，也可知 $t$ 必须为整数。因为 $r>0$ 且 $t>0$，所以 $0 < t = r(1002-r) < 501^2 = 251001$，端点不能取到，因为根必须互不相同且为正。又因为 $r$ 与 $1002-r$ 都不能是整数，所以 $n=-1002t$ 的可能取值个数为 $251,000 - 500 = \boxed{\textbf{(C) } 250,500}$。

Topics