Difference between revisions of "2010 AIME II Problems/Problem 6"
(→Solution) |
|||
Line 15: | Line 15: | ||
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 == |
Latest 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
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.