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

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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
 
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>
 
<cmath> 4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = 528. </cmath>
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== See also ==
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{{AMC12 box|year=2013|ab=B|num-b=24|after=Last Question}}

Revision as of 18:17, 22 February 2013

Problem

Let $G$ be the set of polynomials of the form \[P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,\] where $c_1,c_2,\cdots, c_{n-1}$ are integers and $P(z)$ has distinct roots of the form $a+ib$ with $a$ and $b$ integers. How many polynomials are in $G$?

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

Solution

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

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

We do casework on the (unique) even constant term $d\in\{\pm2,\pm10,\pm50\}$ in our product. For convenience, let $F(d)$ be the number of ways to get a product of $50/d$ without using $\pm 1$ (so only using $\pm5,\pm25$) and recall $f(-1) = 1$; then our final answer will be $2^{f(1)}\sum_{d\in\{2,10,50\}}(f(-d)+f(d))(F(-d)+F(d))$. It's easy to compute $F(-50)=0$, $F(50)=1$, $F(-10)=f(5)=5$, $F(10)=f(-5)=1$, $F(-2)=f(-25)+f(-5)f(5)=6$, $F(2)=f(25)+\binom{f(5)}{2}=16$, so we get \[4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = 528.\]

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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Problem 24
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