Difference between revisions of "2010 AIME II Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | You can | + | You can factor the polynomial into two quadratic factors or a linear and a cubic factor. |
− | + | For two quadratic factors, let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics, so that | |
<cmath>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.</cmath> | <cmath>(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.</cmath> | ||
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Therefore, again setting coefficients equal, <math>a + c = 0\Longrightarrow a=-c</math>, <math>b + d + ac = 0\Longrightarrow b+d=a^2</math> , <math>ad + bc = - n</math>, and so <math>bd = 63</math>. | Therefore, again setting coefficients equal, <math>a + c = 0\Longrightarrow a=-c</math>, <math>b + d + ac = 0\Longrightarrow b+d=a^2</math> , <math>ad + bc = - n</math>, and so <math>bd = 63</math>. | ||
− | Since <math>b+d=a^2</math>, the only possible values for <math>(b,d)</math> are <math>(1,63)</math> and <math>(7,9)</math>. From this we find that the possible values for <math>n</math> are <math>\pm 8 \cdot 62</math> and <math>\pm 4 \cdot 2</math>. Therefore, the answer is <math>\boxed{008}</math>. | + | Since <math>b+d=a^2</math>, the only possible values for <math>(b,d)</math> are <math>(1,63)</math> and <math>(7,9)</math>. From this we find that the possible values for <math>n</math> are <math>\pm 8 \cdot 62</math> and <math>\pm 4 \cdot 2</math>. |
+ | |||
+ | For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest <math>n</math> in that case to be <math>48</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{008}</math>. | ||
== See also == | == See also == |
Revision as of 14:06, 30 July 2017
Problem
Find the smallest positive integer with the property that the polynomial can be written as a product of two nonconstant polynomials with integer coefficients.
Solution
You can factor the polynomial into two quadratic factors or a linear and a cubic factor.
For two quadratic factors, let and be the two quadratics, so that
Therefore, again setting coefficients equal, , , , and so .
Since , the only possible values for are and . From this we find that the possible values for are and .
For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest in that case to be .
Therefore, the answer is .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.