Difference between revisions of "2010 AIME II Problems/Problem 10"
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| − | == Problem | + | == Problem == |
| − | Find the number of second-degree | + | Find the number of second-degree [[polynomial]]s <math>f(x)</math> with integer [[coefficient]]s and integer zeros for which <math>f(0)=2010</math>. |
| + | __TOC__ | ||
==Solution== | ==Solution== | ||
===Solution 1=== | ===Solution 1=== | ||
| − | Let <math>f(x) = a(x-r)(x-s)</math>. Then <math>ars=2010=2\cdot3\cdot5\cdot67</math>. First consider the case where <math>r</math> and <math>s</math> (and thus <math>a</math>) are positive. | + | Let <math>f(x) = a(x-r)(x-s)</math>. Then <math>ars=2010=2\cdot3\cdot5\cdot67</math>. First consider the case where <math>r</math> and <math>s</math> (and thus <math>a</math>) are positive. There are <math>3^4 = 81</math> ways to split up the prime factors between <math>a</math>, <math>r</math>, and <math>s</math>. However, <math>r</math> and <math>s</math> are indistinguishable. In one case, <math>(a,r,s) = (2010,1,1)</math>, we have <math>r=s</math>. The other <math>80</math> cases are double counting, so there are <math>40</math>. |
| − | There are <math>3^4 = 81</math> ways to split up the prime factors between a, r, and s. However, r and s are indistinguishable. In one case, <math>(a,r,s) = (2010,1,1)</math>, we have <math>r=s</math>. The other <math>80</math> cases are double counting, so there are <math>40</math>. | ||
We must now consider the various cases of signs. For the <math>40</math> cases where <math>|r|\neq |s|</math>, there are a total of four possibilities, For the case <math>|r|=|s|=1</math>, there are only three possibilities, <math>(r,s) = (1,1); (1,-1); (-1,-1)</math> as <math>(-1,1)</math> is not distinguishable from the second of those three. | We must now consider the various cases of signs. For the <math>40</math> cases where <math>|r|\neq |s|</math>, there are a total of four possibilities, For the case <math>|r|=|s|=1</math>, there are only three possibilities, <math>(r,s) = (1,1); (1,-1); (-1,-1)</math> as <math>(-1,1)</math> is not distinguishable from the second of those three. | ||
| − | Thus the grand total is <math>4\cdot40 + 3 = \boxed{163}</math> | + | |
| + | Thus the grand total is <math>4\cdot40 + 3 = \boxed{163}</math>. | ||
===Solution 2=== | ===Solution 2=== | ||
| − | + | Due to [[Burnside's Lemma]], the answer is <math>\frac{3^4+1}{2}+\frac{3^4+1}{2}+3^4=\boxed{163}</math> — it is group action of <math>\mathbb{Z}^2</math>. | |
| − | Burnside's Lemma | ||
| − | |||
| − | |||
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=9|num-a=11|n=II}} | {{AIME box|year=2010|num-b=9|num-a=11|n=II}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 10:58, 6 April 2010
Problem
Find the number of second-degree polynomials 
with integer coefficients and integer zeros for which 
.
Solution
Solution 1
Let 
. Then 
. First consider the case where 
and 
(and thus 
) are positive. There are 
ways to split up the prime factors between , , and . However, and are indistinguishable. In one case, 
, we have 
. The other 
cases are double counting, so there are 
.
We must now consider the various cases of signs. For the cases where 
, there are a total of four possibilities, For the case 
, there are only three possibilities, 
as 
is not distinguishable from the second of those three.
Thus the grand total is 
.
Solution 2
Due to Burnside's Lemma, the answer is 
— it is group action of 
.
See also
| 2010 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
