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. | |
− | < | + | For two quadratic factors, let <math>x^2+ax+b</math> and <math>x^2+cx+d</math> be the two quadratics, so that |
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− | + | <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> | |
− | + | 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>. | |
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− | < | + | 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>. | |
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− | + | 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 |
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