Difference between revisions of "2010 AIME II Problems/Problem 6"

(Remove extra problem section)
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 <math>\boxed{008}</math>.
+
Therefore, the answer is 4*2 = <math>\boxed{008}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:17, 9 May 2020

Problem

Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ 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 $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that

\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\]

Therefore, again setting coefficients equal, $a + c = 0\Longrightarrow a=-c$, $b + d + ac = 0\Longrightarrow b+d=a^2$ , $ad + bc = - n$, and so $bd = 63$.

Since $b+d=a^2$, the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$. From this we find that the possible values for $n$ are $\pm 8 \cdot 62$ and $\pm 4 \cdot 2$.

For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest $n$ in that case to be $48$.

Therefore, the answer is 4*2 = $\boxed{008}$.

See also

2010 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png