Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
(→‎Solution 2)
 
(3 intermediate revisions by 3 users not shown)
Line 18: Line 18:
  
 
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>.
 
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 ==
 
== Solution 2 ==
  
−
Letting the roots be <math>r</math>, <math>s</math>, and <math>t</math>, where <math>t = r+s</math>, we see that by Vieta's Formula's, <math>2004 = r+s+t = t + t = 2t</math>, and so <math>t = 1002</math>. Therefore, <math>x-1002</math> is a factor of <math>x^3 - 2004x^2 + mx + n</math>. Letting <math>x = 0</math> gives that <math>1002 \mid n</math> because <math>x - 1002 \mid x^3 - 2004x^2 + mx + n</math>. Letting <math>n = 1002a</math> and noting that <math>x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)</math> for some <math>b</math>, we see that <math>b</math> is the sum of the roots of <math>x^2 - bx + a</math>, <math>r</math> and <math>s</math>, and so <math>b = 1002</math>. Now, we have that <math>x^2 - 1002x + a</math> has roots <math>r</math> and <math>s</math>, and we wish to find the number of possible values of <math>a</math>. By the quadratic formula, we see that <cmath>\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}</cmath> are the two values of noninteger positive real numbers <math>r</math> and <math>s</math>, neither of which is equal to <math>1002</math>. This information gives us that <math>0 < 501^2 - a < 501^2</math>, and so since <math>501^2 - a</math> is evidently not a square, we have <math>501^2 - 1 - 500 = 251,001 - 500 - 1 = \boxed{\textbf{(C) } 250,500}</math> possible values of <math>n = 1002a</math>.
+
Letting the roots be <math>r</math>, <math>s</math>, and <math>t</math>, where <math>t = r+s</math>, we see that by Vieta's Formula's, <math>2004 = r+s+t = t + t = 2t</math>, and so <math>t = 1002</math>. Therefore, <math>x-1002</math> is a factor of <math>x^3 - 2004x^2 + mx + n</math>. Letting <math>x = 0</math> gives that <math>1002 \mid n</math> because <math>x - 1002 \mid x^3 - 2004x^2 + mx + n</math>. Letting <math>n = -1002a</math> and noting that <math>x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)</math> for some <math>b</math>, we see that <math>b</math> is the sum of the roots of <math>x^2 - bx + a</math>, <math>r</math> and <math>s</math>, and so <math>b = 1002</math>. Now, we have that <math>x^2 - 1002x + a</math> has roots <math>r</math> and <math>s</math>, and we wish to find the number of possible values of <math>a</math>. By the quadratic formula, we see that <cmath>\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}</cmath> are the two values of noninteger positive real numbers <math>r</math> and <math>s</math>, neither of which is equal to <math>1002</math>. This information gives us that <math>0 < 501^2 - a < 501^2</math>, and so since <math>501^2 - a</math> is evidently not a square, we have <math>501^2 - 1 - 500 = 251,001 - 500 - 1 = \boxed{\textbf{(C) } 250,500}</math> possible values of <math>n = 1002a</math>.
 +
 
 +
== Solution 3 (cheese)==
 +
 
 +
Observe that the answer clearly must have something to do with the number <math>2004</math>, and we see that <math>250,500</math> is a multiple of <math>2004</math>, so there is a very high probability that it is the correct answer.
  
 
== See also ==
 
== See also ==

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_23&oldid=200454"