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

Revision as of 09:32, 30 December 2022

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 1

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

Solution 2

Disregard sign; we can tack on $x-1$ if the product ends up being negative.

$1: \pm i,-1$ (2) (1 is not included)

$2: \pm 2, \pm 1\pm i$ (4)

$5: \pm 2\pm i, \pm 1\pm 2i, \pm 5$ (6)

$10: \pm 3\pm i, \pm 1\pm 3i, \pm 10$ (6)

$25: \pm 25, \pm 3\pm 4i, \pm 4\pm 3i, \pm 5i$ (7)

$50: \pm 50, \pm 1\pm 7i, \pm7\pm i, \pm 5\pm 5i$ (8)

Our answer is $2^2\left(4\cdot\binom{6}{2}+6\cdot 6+4\cdot 7+8\right)=\boxed{528.}$

Solution 3

By Vieta's formula $50$ is the product of all $n$ roots. As the roots are all in the form $a + bi$ , there must exist a conjugate $a-bi$ for each root.

$(a+bi)(a-bi) = a^2 + b^2$

$50 = 2 \cdot 5^2$

If $a \neq b \neq 0$ , the root can be $a \pm bi$ , $-a \pm bi$ , $b \pm ai$ , $-b \pm ai$ , totaling $8$ pairs of roots.

If $a = b$ , the root can be $a \pm ai$ , $-a \pm ai$ , totaling $2$ pairs of roots.

If $a = 0$ , $a \neq b$ , the root can be $\pm a$ , $\pm ai$ , totaling $2$ pairs of roots.

$\begin{align*} (1^2+1^2)(3^2+4^2) &: 4 \cdot 8 = 32\\ (1^2+1^2)5^2 &: 4 \cdot 4 = 16\\ (1^2+3^2)(1^2+2^2) &: 8 \cdot 8 = 64\\ (1^2+1^2)(1^2+2^2)(1^2+2^2) &: 4 \cdot 8 \cdot 8 = 256\\ (1^2+7^2) &: 8 \end{align*}$

@@ Line 7: / Line 7: @@
 <math> \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056 </math>
-==Solution==
+==Solution 1==
-===Solution 1===
 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>.
@@ Line 16: / Line 15: @@
 <cmath> 4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = \boxed{\textbf{(B) }528} </cmath>
-===Solution 2===
+==Solution 2==
 Disregard sign; we can tack on <math>x-1</math> if the product ends up being negative.
@@ Line 33: / Line 32: @@
 Our answer is <math>2^2\left(4\cdot\binom{6}{2}+6\cdot 6+4\cdot 7+8\right)=\boxed{528.}</math>
+==Solution 3==
+By Vieta's formula <math>50</math> is the product of all <math>n</math> roots. As the roots are all in the form <math>a + bi</math>, there must exist a conjugate <math>a-bi</math> for each root.
+<math>(a+bi)(a-bi) = a^2 + b^2</math>
+<math>50 = 2 \cdot 5^2</math>
+If <math>a \neq b \neq 0</math>, the root can be <math>a \pm bi</math>, <math>-a \pm bi</math>, <math>b \pm ai</math>, <math>-b \pm ai</math>, totaling <math>8</math> pairs of roots.
+If <math>a = b</math>, the root can be <math>a \pm ai</math>, <math>-a \pm ai</math>, totaling <math>2</math> pairs of roots.
+If <math>a = 0</math>, <math>a \neq b</math>, the root can be <math>\pm a</math>, <math>\pm ai</math>, totaling <math>2</math> pairs of roots.
+<cmath>\begin{align*}
+(1^2+1^2)(3^2+4^2) &: 4 \cdot 8 = 32\\
+(1^2+1^2)5^2 &: 4 \cdot 4 = 16\\
+(1^2+3^2)(1^2+2^2) &: 8 \cdot 8 = 64\\
+(1^2+1^2)(1^2+2^2)(1^2+2^2) &: 4 \cdot 8 \cdot 8 = 256\\
+(1^2+7^2) &: 8
+\end{align*}</cmath>
 == See also ==

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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