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Difference between revisions of "2004 AMC 12B Problems/Problem 23"
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\qquad\mathrm{(D)}\ 250,750
 
\qquad\mathrm{(D)}\ 250,750
 
\qquad\mathrm{(E)}\ 251,000</math>
 
\qquad\mathrm{(E)}\ 251,000</math>
 +
 
== Solution ==
 
== Solution ==
 
Let the roots be <math>r,s,r + s</math>, and let <math>t = rs</math>. Then
 
Let the roots be <math>r,s,r + s</math>, and let <math>t = rs</math>. Then
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and we need the number of possible products <math>t = rs = r(1002 - r)</math>.  
 
and we need the number of possible products <math>t = rs = r(1002 - r)</math>.  
  
−
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. Because <math>r</math> cannot be an integer, there are <math>251000 - 500 = 250,500 \mathrm{(C)}</math> possible values of <math>n = -1002t</math>.
+
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. Because <math>r</math> cannot be an integer, there are <math>251000 - 500 = 250,500\ \mathrm{(C)}</math> possible values of <math>n = -1002t</math>.
  
 
== See also ==
 
== See also ==
−
{{AMC12 box|year=2004|ab=B|num-b=23|num-a=25}}
+
{{AMC12 box|year=2004|ab=B|num-b=22|num-a=24}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 10:07, 10 February 2008

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 $n = -1002t$.

See also

2004 AMC 12B (Problems • Answer Key • Resources)
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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