2013 AMC 12B Problems/Problem 25

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 $G$ ?

$\textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056$

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 $4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = \boxed{\textbf{(B) }528}$

2013 AMC 12B (Problems • Answer Key • Resources)
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2013 AMC 12B Problems/Problem 25

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