Difference between revisions of "2004 AMC 12B Problems/Problem 23"

Latest revision as of 13:52, 29 October 2023

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

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 $501^2 - 1 - 500 = 251,001 - 500 - 1 = \boxed{\textbf{(C) } 250,500}$ possible values of $n = 1002a$ .

Solution 3 (cheese)

Observe that the answer clearly must have something to do with the number $2004$ , and we see that $250,500$ is a multiple of $2004$ , so there is a very high probability that it is the correct answer.

@@ Line 7: / Line 7: @@
 \qquad\mathrm{(D)}\ 250,\!750
 \qquad\mathrm{(E)}\ 251,\!000</math>
+== Solution 1 ==
+Let the roots be <math>r,s,r + s</math>, and let <math>t = rs</math>. Then
+<center><math>(x - r)(x - s)(x - (r + s))</math> <math>= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0</math></center>
+and by matching coefficients, <math>2(r + s) = 2004 \Longrightarrow r + s = 1002</math>. Then our polynomial looks like
+<cmath>x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0</cmath>
+and we need the number of possible products <math>t = rs = r(1002 - r)</math>. Because <math>m=t+1002^2</math> is an integer, we also note that <math>t</math> must be an integer.
+Since <math>r > 0</math> and <math>t > 0</math>, it follows that <math>0 < t = r(1002-r) < 501^2 = 251001</math>, with the endpoints not achievable because the roots must be distinct and positive. Because neither <math>r</math> nor <math>1002-r</math> can be an integer, there are <math>251,000 - 500 = \boxed{\textbf{(C) } 250,500}</math> possible values of <math>n = -1002t</math>.
 == Solution 2 ==

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