Difference between revisions of "2004 AMC 12B Problems/Problem 23"
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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>. | ||
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== 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)== | == Solution 3 (cheese)== |
Latest revision as of 13:52, 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
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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