# Difference between revisions of "2013 AMC 12B Problems/Problem 25"

(Created page with "==Problem== Let <math>G</math> be the set of polynomials of the form <cmath> P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50, </cmath> where <math> c_1,c_2,\cdots, c_{n-1} </math>...") |
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<math> \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D}}\ 992\qquad\textbf{(E)}\ 1056 </math> | <math> \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D}}\ 992\qquad\textbf{(E)}\ 1056 </math> | ||

+ | ==Solution== | ||

+ | If we factor into irreducible polynomials (in <math>\mathbb{Q}[x]</math>), each factor <math>f_i</math> has exponent <math>1</math> in the factorization and degree at most <math>2</math> (since the <math>a+bi</math> with <math>b\ne0</math> come in conjugate pairs with product <math>a^2+b^2</math>). Clearly we want the product of constant terms of these polynomials to equal <math>50</math>; for <math>d\mid 50</math>, let <math>f(d)</math> be the number of permitted <math>f_i</math> with constant term <math>d</math>. It's easy to compute <math>f(1)=2</math>, <math>f(2)=3</math>, <math>f(5)=5</math>, <math>f(10)=5</math>, <math>f(25)=6</math>, <math>f(50)=7</math>, and obviously <math>f(d) = 1</math> for negative <math>d\mid 50</math>. | ||

+ | |||

+ | Note that by the distinctness condition, the only constant terms <math>d</math> that can be repeated are those with <math>d^2\mid 50</math> and <math>f(d)>1</math>, i.e. <math>+1</math> and <math>+5</math>. Also, the <math>+1</math>s don't affect the product, so we can simply count the number of polynomials with no constant terms of <math>+1</math> and multiply by <math>2^{f(1)} = 4</math> at the end. | ||

+ | |||

+ | We do casework on the (unique) even constant term <math>d\in\{\pm2,\pm10,\pm50\}</math> in our product. For convenience, let <math>F(d)</math> be the number of ways to get a product of <math>50/d</math> without using <math>\pm 1</math> (so only using <math>\pm5,\pm25</math>) and recall <math>f(-1) = 1</math>; then our final answer will be <math>2^{f(1)}\sum_{d\in\{2,10,50\}}(f(-d)+f(d))(F(-d)+F(d))</math>. It's easy to compute <math>F(-50)=0</math>, <math>F(50)=1</math>, <math>F(-10)=f(5)=5</math>, <math>F(10)=f(-5)=1</math>, <math>F(-2)=f(-25)+f(-5)f(5)=6</math>, <math>F(2)=f(25)+\binom{f(5)}{2}=16</math>, so we get | ||

+ | <cmath> 4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = 528. </cmath> |

## Revision as of 16:55, 22 February 2013

## Problem

Let be the set of polynomials of the form where are integers and has distinct roots of the form with and integers. How many polynomials are in ?

$\textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D}}\ 992\qquad\textbf{(E)}\ 1056$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

## Solution

If we factor into irreducible polynomials (in ), each factor has exponent in the factorization and degree at most (since the with come in conjugate pairs with product ). Clearly we want the product of constant terms of these polynomials to equal ; for , let be the number of permitted with constant term . It's easy to compute , , , , , , and obviously for negative .

Note that by the distinctness condition, the only constant terms that can be repeated are those with and , i.e. and . Also, the s don't affect the product, so we can simply count the number of polynomials with no constant terms of and multiply by at the end.

We do casework on the (unique) even constant term in our product. For convenience, let be the number of ways to get a product of without using (so only using ) and recall ; then our final answer will be . It's easy to compute , , , , , , so we get