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

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 $n = -1002t$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2004 AMC 12B Problems/Problem 23

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