Difference between revisions of "2004 AMC 12B Problems/Problem 23"
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The [[polynomial]] <math>x^3 - 2004 x^2 + mx + n</math> 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 <math>n</math> are possible? | The [[polynomial]] <math>x^3 - 2004 x^2 + mx + n</math> 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 <math>n</math> are possible? | ||
− | <math>\mathrm{(A)}\ 250,000 | + | <math>\mathrm{(A)}\ 250,\!000 |
− | \qquad\mathrm{(B)}\ 250,250 | + | \qquad\mathrm{(B)}\ 250,\!250 |
− | \qquad\mathrm{(C)}\ 250,500 | + | \qquad\mathrm{(C)}\ 250,\!500 |
− | \qquad\mathrm{(D)}\ 250,750 | + | \qquad\mathrm{(D)}\ 250,\!750 |
− | \qquad\mathrm{(E)}\ 251,000</math> | + | \qquad\mathrm{(E)}\ 251,\!000</math> |
− | == Solution == | + | == Solution 1 == |
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 by matching coefficients, <math>2(r + s) = 2004 \Longrightarrow r + s = 1002</math>. Then our polynomial looks like | 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> | <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>. | + | 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. Because <math>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 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>. |
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+ | == 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>. | ||
+ | |||
+ | == 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 == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:52, 29 October 2023
Problem
The polynomial 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
are possible?
Solution 1
Let the roots be , and let
. Then
![$(x - r)(x - s)(x - (r + s))$](http://latex.artofproblemsolving.com/9/a/a/9aa6b4356e1b571b8c6b0273d4a5ee73ddd93031.png)
![$= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0$](http://latex.artofproblemsolving.com/1/8/f/18f8cb41503afe71a27d8834fd83f6ee33165a95.png)
and by matching coefficients, . Then our polynomial looks like
and we need the number of possible products
. Because
is an integer, we also note that
must be an integer.
Since and
, it follows that
, with the endpoints not achievable because the roots must be distinct and positive. Because neither
nor
can be an integer, there are
possible values of
.
Solution 2
Letting the roots be ,
, and
, where
, we see that by Vieta's Formula's,
, and so
. Therefore,
is a factor of
. Letting
gives that
because
. Letting
and noting that
for some
, we see that
is the sum of the roots of
,
and
, and so
. Now, we have that
has roots
and
, and we wish to find the number of possible values of
. By the quadratic formula, we see that
are the two values of noninteger positive real numbers
and
, neither of which is equal to
. This information gives us that
, and so since
is evidently not a square, we have
possible values of
.
Solution 3 (cheese)
Observe that the answer clearly must have something to do with the number , and we see that
is a multiple of
, 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 |
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