2004 AMC 12B Problems/Problem 23

Problem

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?

$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,\!750 \qquad\mathrm{(E)}\ 251,\!000$

Solution 1

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 $t = rs = r(1002 - r)$.

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. Because $r$ cannot be an integer, there are $251000 - 500 = 250,\!500\ \mathrm{(C)}$ possible values of $n = -1002t$.

Solution 2

Letting the roots be $r$, $s$, and $t$, where $t = r+s$, we see that by Vieta's Formula's, $2004 = r+s+t = t + t = 2t$, and so $t = 1002$. Therefore, $x-1002$ is a factor of $x^3 - 2004x^2 + mx + n$. Letting $x = 0$ gives that $1002 \mid n$ because $x - 1002 \mid x^3 - 2004x^2 + mx + n$. Letting $n = 1002a$ and noting that $x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)$ for some $b$, we see that $b$ is the sum of the roots of $x^2 - bx + a$, $r$ and $s$, and so $b = 1002$. Now, we have that $x^2 - 1002x + a$ has roots $r$ and $s$, and we wish to find the number of possible values of $a$. By the quadratic formula, we see that \[\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}\] are the two values of noninteger positive real numbers $r$ and $s$, neither of which is equal to $1002$. This information gives us that $0 < 501^2 - a < 501^2$, and so since $501^2 - a$ is evidently not a square, we have $501^2 - 1 - 500 = 251,001 - 500 - 1 = 250,\!500\ \mathrm{(C)}$ possible values of $n = 1002a$.

See also

2004 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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