Difference between revisions of "2010 AIME II Problems/Problem 6"
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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>. | 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 4 | + | Therefore, the answer is <math>4 \cdot 2 = \boxed{008}</math>. |
== See also == | == See also == | ||
Revision as of 22:42, 28 May 2021
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
![\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\]](http://latex.artofproblemsolving.com/1/e/2/1e2074dedb9b951ecafd50176871218ee0fdf3e5.png)
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. 
