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

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== Problem 10 ==
 
== Problem 10 ==
How many ordered pairs of positive integers (M,N) satisfy the equation <math>\frac {M}{6}</math>    =    <math>\frac{6}{N}</math>
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How many ordered pairs of positive integers <math>(M,N)</math> satisfy the equation <math>\frac {M}{6}</math>    =    <math>\frac{6}{N}</math>?
  
 
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math>
 
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math>

Revision as of 14:19, 28 August 2020

Problem 10

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

Solution

$\frac {M}{6}$ = $\frac{6}{N}$

is a ratio; therefore, you can cross-multiply.

$MN=36$

Now you find all the factors of 36:

$1\times36=36$

$2\times18=36$

$3\times12=36$

$4\times9=36$

$6\times6=36$.

Now you can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order.

$4\cdot 2+1=9$

$\boxed{\textbf{(D)}\ 9}$

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
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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