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

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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>
  
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==Solution==
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==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>.
 
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>.
  
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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] = \boxed{\textbf{(B) }528} </cmath>
 
<cmath> 4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = \boxed{\textbf{(B) }528} </cmath>
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==Solution 2==
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Disregard sign; we can tack on <math>x-1</math> if the product ends up being negative.
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<math>1: \pm i,-1</math> (2) (1 is not included)
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<math>2: \pm 2, \pm 1\pm i</math> (4)
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<math>5: \pm 2\pm i, \pm 1\pm 2i, \pm 5</math> (6)
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<math>10: \pm 3\pm i, \pm 1\pm 3i, \pm 10</math> (6)
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<math>25: \pm 25, \pm 3\pm 4i, \pm 4\pm 3i, \pm 5i</math> (7)
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<math>50: \pm 50, \pm 1\pm 7i, \pm7\pm i, \pm 5\pm 5i</math> (8)
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Our answer is <math>2^2\left(4\cdot\binom{6}{2}+6\cdot 6+4\cdot 7+8\right)=\boxed{528.}</math>
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==Solution 3==
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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.
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<math>(a+bi)(a-bi) = a^2 + b^2</math>
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<math>50 = 2 \cdot 5^2</math>
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If <math>a \neq b \neq 0</math>, the roots can be <math>a \pm bi</math>, <math>-a \pm bi</math>, <math>b \pm ai</math>, <math>-b \pm ai</math>, totaling <math>4</math> pairs of roots.
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If <math>a = b</math>, the roots can be <math>a \pm ai</math>, <math>-a \pm ai</math>, totaling <math>2</math> pairs of roots.
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If <math>a \neq b</math>, <math>b = 0</math>, the roots can be <math>\pm a</math>, <math>\pm ai</math>, totaling <math>2</math> pairs of roots.
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<cmath>\begin{align*}
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2 \cdot 25 &= (1^2+1^2)5^2 &: 2 \cdot 2 = 4\\
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2 \cdot 25 &= 2 \cdot 5^2 &: 2 \cdot 2 = 4\\
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2 \cdot 25 &= (1^2+1^2) \cdot (3^2+4^2) &: 2 \cdot 4 = 8\\
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2 \cdot 25 &= 2 \cdot (3^2+4^2) &: 2 \cdot 4 = 8
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\end{align*}</cmath>
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<cmath>\begin{align*}
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10 \cdot 5 &= (1^2+3^2)(1^2+2^2) &&: 4 \cdot 4 = 16\\
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10 \cdot 5 &= 10 \cdot (1^2+2^2) &&: 2 \cdot 4 = 8\\
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10 \cdot 5 &= (1^2+3^2) \cdot 5 &&: 4 \cdot 2 = 8\\
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10 \cdot 5 &= 10 \cdot 5 &&: 2 \cdot 2 = 4\\
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\end{align*}</cmath>
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<cmath>\begin{align*}
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2 \cdot 5  \cdot 5&= (1^2+1^2)(1^2+2^2)(1^2+2^2) &&: 2 \cdot 4 \cdot 4 = 32\\
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2 \cdot 5  \cdot 5&= 2 \cdot (1^2+2^2)(1^2+2^2) &&: 2 \cdot 4 \cdot 4 = 32\\
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2 \cdot 5  \cdot 5&= 2 \cdot 5 \cdot (1^2+2^2) &&: 2 \cdot 2 \cdot 4 = 16\\
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2 \cdot 5  \cdot 5&= 2 \cdot 5 \cdot 5 &&: 2 \cdot 2 \cdot 2 = 8\\
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2 \cdot 5  \cdot 5&= (1^2+1^2) \cdot 5 \cdot (1^2+2^2) &&: 2 \cdot 2 \cdot 4 = 16\\
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2 \cdot 5  \cdot 5&= (1^2+1^2) \cdot 5 \cdot 5 &&: 2 \cdot 2 \cdot 2 = 8\\
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\end{align*}</cmath>
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<cmath>\begin{align*}
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(1^2+7^2) &: 4\\
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(5^2+5^2) &: 2\\
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50 &: 2
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\end{align*}</cmath>
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<math>4+4+8+8+16+8+8+4+32+32+16+8+16+8+4+2+2 = 176</math>
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For each case <math>1^2</math> can be added, yielding 2 more cases <math>(\pm 1, \pm i)</math>. <math>176 \cdot 3 = \boxed{\textbf{(B) }528}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
== See also ==
 
== See also ==

Latest revision as of 00:10, 31 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 roots can be $a \pm bi$, $-a \pm bi$, $b \pm ai$, $-b \pm ai$, totaling $4$ pairs of roots.

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

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

\begin{align*} 2 \cdot 25 &= (1^2+1^2)5^2 &: 2 \cdot 2 = 4\\ 2 \cdot 25 &= 2 \cdot 5^2 &: 2 \cdot 2 = 4\\ 2 \cdot 25 &= (1^2+1^2) \cdot (3^2+4^2) &: 2 \cdot 4 = 8\\ 2 \cdot 25 &= 2 \cdot (3^2+4^2) &: 2 \cdot 4 = 8 \end{align*}

\begin{align*} 10 \cdot 5 &= (1^2+3^2)(1^2+2^2) &&: 4 \cdot 4 = 16\\ 10 \cdot 5 &= 10 \cdot (1^2+2^2) &&: 2 \cdot 4 = 8\\ 10 \cdot 5 &= (1^2+3^2) \cdot 5 &&: 4 \cdot 2 = 8\\ 10 \cdot 5 &= 10 \cdot 5 &&: 2 \cdot 2 = 4\\ \end{align*}

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

\begin{align*} (1^2+7^2) &: 4\\ (5^2+5^2) &: 2\\ 50 &: 2 \end{align*}

$4+4+8+8+16+8+8+4+32+32+16+8+16+8+4+2+2 = 176$

For each case $1^2$ can be added, yielding 2 more cases $(\pm 1, \pm i)$. $176 \cdot 3 = \boxed{\textbf{(B) }528}$

~isabelchen

See also

2013 AMC 12B (Problems • Answer Key • Resources)
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