Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 3"
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==Solution== | ==Solution== | ||
+ | :''The following solution is non-rigorous''. | ||
− | {{ | + | We would normally expect 2007 terms after multiplying out all of the binomials, but our goal is minimize the number of non-zero terms. We could get rid of some terms by applying repeated [[difference of squares]]. In other words, we let <math>r_1 = -r_2, \ldots, r_{2005} = -r_{2006}</math>. Then our polynomial reduces to <math>(x^2 - r_1^2)(x^2 - r_3^2)\cdots (x^2 - r_{2005}^2)</math>. This is the product of <math>1003</math> binomials, which gives us <math>1004</math> terms (with nonzero coefficients). Since <math>1004 = 2^2 \cdot 251</math> (assuming the constant term counts as a coefficient), our answer is <math>251</math>. |
+ | Note that we could not apply difference of cubes etc, since that would require complex roots. | ||
− | --- | + | ==See also== |
+ | {{Mock AIME box|year=2006-2007|n=4|num-b=2|num-a=4|source=125025}} | ||
− | + | [[Category:Intermediate Algebra Problems]] | |
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Latest revision as of 14:44, 8 October 2007
Problem
Find the largest prime factor of the smallest positive integer such that are distinct integers such that the polynomial has exactly nonzero coefficients.
Solution
- The following solution is non-rigorous.
We would normally expect 2007 terms after multiplying out all of the binomials, but our goal is minimize the number of non-zero terms. We could get rid of some terms by applying repeated difference of squares. In other words, we let . Then our polynomial reduces to . This is the product of binomials, which gives us terms (with nonzero coefficients). Since (assuming the constant term counts as a coefficient), our answer is .
Note that we could not apply difference of cubes etc, since that would require complex roots.
See also
Mock AIME 4 2006-2007 (Problems, Source) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |