Difference between revisions of "2013 AMC 12B Problems/Problem 25"
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where <math> c_1,c_2,\cdots, c_{n-1} </math> are integers and <math>P(z)</math> has distinct roots of the form <math>a+ib</math> with <math>a</math> and <math>b</math> integers. How many polynomials are in <math>G</math>? | where <math> c_1,c_2,\cdots, c_{n-1} </math> are integers and <math>P(z)</math> has distinct roots of the form <math>a+ib</math> with <math>a</math> and <math>b</math> integers. How many polynomials are in <math>G</math>? | ||
− | <math> \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D | + | <math> \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056 </math> |
+ | |||
==Solution== | ==Solution== | ||
+ | ===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] = 528. </ | + | <cmath> 4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = \boxed{\textbf{(B) }528} </cmath> |
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | Disregard sign; we can tack on <math>x-1</math> if the product ends up being negative. | ||
+ | |||
+ | <math>1: \pm i,-1</math> (2) (1 is not included) | ||
+ | |||
+ | <math>2: \pm 2, \pm 1\pm i</math> (4) | ||
+ | |||
+ | <math>5: \pm 2\pm i, \pm 1\pm 2i, \pm 5</math> (6) | ||
+ | |||
+ | <math>10: \pm 3\pm i, \pm 1\pm 3i, \pm 10</math> (6) | ||
+ | |||
+ | <math>25: \pm 25, \pm 3\pm 4i, \pm 4\pm 3i, \pm 5i</math> (7) | ||
+ | |||
+ | <math>50: \pm 50, \pm 1\pm 7i, \pm7\pm i, \pm 5\pm 5i</math> (8) | ||
+ | |||
+ | Our answer is <math>2^2\left(4\cdot\binom{6}{2}+6\cdot 6+4\cdot 7+8\right)=\boxed{528.}</math> | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=B|num-b=24|after=Last Question}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 20:20, 28 December 2017
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 ?
Solution
Solution 1
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
Solution 2
Disregard sign; we can tack on if the product ends up being negative.
(2) (1 is not included)
(4)
(6)
(6)
(7)
(8)
Our answer is
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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