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Difference between revisions of "2012 AMC 10B Problems/Problem 10"

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== Problem 10 ==
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== Problem ==
 
How many ordered pairs of positive integers <math>(M,N)</math> satisfy the equation <math>\frac{M}{6}=\frac{6}{N}?</math>
 
How many ordered pairs of positive integers <math>(M,N)</math> satisfy the equation <math>\frac{M}{6}=\frac{6}{N}?</math>
  
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~MRENTHUSIASM (Reformatting)
 
~MRENTHUSIASM (Reformatting)
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== Solution 2 ==
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Cross-multiplying gives <math>MN=36.</math> From the prime factorization <cmath>36=2^2\cdot3^2,</cmath> we conclude that <math>36</math> has <math>(2+1)(2+1)=9</math> positive divisors. There are <math>9</math> values of <math>M,</math> and each value generates <math>1</math> ordered pair <math>(M,N).</math> So, there are <math>\boxed{\textbf{(D)}\ 9}</math> ordered pairs <math>(M,N)</math> in total.
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~MRENTHUSIASM
  
 
==See Also==
 
==See Also==

Latest revision as of 22:08, 3 September 2021

Problem

How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution 1

Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers: \begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*} The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the factors. The product $6\cdot6$ produces $1$ ordered pair $(M,N).$ Together, we have $4\cdot2+1=\boxed{\textbf{(D)}\ 9}$ ordered pairs $(M,N).$

~Rguan (Solution)

~MRENTHUSIASM (Reformatting)

Solution 2

Cross-multiplying gives $MN=36.$ From the prime factorization \[36=2^2\cdot3^2,\] we conclude that $36$ has $(2+1)(2+1)=9$ positive divisors. There are $9$ values of $M,$ and each value generates $1$ ordered pair $(M,N).$ So, there are $\boxed{\textbf{(D)}\ 9}$ ordered pairs $(M,N)$ in total.

~MRENTHUSIASM

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
All AMC 10 Problems and Solutions

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