Difference between revisions of "2010 AIME II Problems/Problem 6"
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Find the smallest positive integer <math>n</math> with the property that the [[polynomial]] <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with integer coefficients. | Find the smallest positive integer <math>n</math> with the property that the [[polynomial]] <math>x^4 - nx + 63</math> can be written as a product of two nonconstant polynomials with integer coefficients. | ||
− | ==Solution== | + | ==Solution 1== |
You can factor the polynomial into two quadratic factors or a linear and a cubic factor. | You can factor the polynomial into two quadratic factors or a linear and a cubic factor. | ||
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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 <math>\boxed{008}</math>. | + | Therefore, the answer is <math>4 \cdot 2 = \boxed{008}</math>. |
+ | == Solution 2 == | ||
+ | Let <math>x^4-nx+63=(x^2+ax+b)(x^2+cx+d)</math>. From this, we get that <math>bd=63\implies d=\frac{63}{b}</math> and <math>a+c=0\implies c=-a</math>. Plugging this back into the equation, we get <math>x^4-nx+63=(x^2+ax+b)\left(x^2-ax+\frac{63}{b}\right)</math>. Expanding gives us <math>x^4-nx+63=x^4+\left(-a^2+b+\frac{63}{b}\right)x^2+\left(\frac{63a}{b}-ab\right)x+63</math>. Therefore <math>-a^2+b+\frac{63}{b}=0</math>. Simplifying gets us <math>b(a^2-b)=63</math>. Since <math>a</math> and <math>b</math> must be integers, we can use guess and check for values of <math>b</math> because <math>b</math> must be a factor of <math>63</math>. Note that <math>b</math> cannot be negative because <math>a</math> would be imaginary. After guessing and checking, we find that the possible values of <math>(a,b)</math> are <math>(\pm 8, 1), (\pm 4, 7), (\pm 4, 9),</math> and <math>(\pm 8, 63)</math>. We have that <math>n=ab-\frac{63a}{b}</math>. Plugging in our values for <math>a</math> and <math>b</math>, we get that the smallest positive integer value <math>n</math> can be is <math>\boxed{008}</math>. | ||
+ | -Heavytoothpaste | ||
== See also == | == See also == |
Latest revision as of 17:47, 29 June 2022
Contents
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 1
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 .
Solution 2
Let . From this, we get that and . Plugging this back into the equation, we get . Expanding gives us . Therefore . Simplifying gets us . Since and must be integers, we can use guess and check for values of because must be a factor of . Note that cannot be negative because would be imaginary. After guessing and checking, we find that the possible values of are and . We have that . Plugging in our values for and , we get that the smallest positive integer value can be is . -Heavytoothpaste
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.