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

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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.
 
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.
  
<math>4*2+1=9</math>  
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<math>4\cdot 2+1=9</math>  
  
 
<math>\boxed{\textbf{(D)}\ 9}</math>
 
<math>\boxed{\textbf{(D)}\ 9}</math>
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==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2012|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2012|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:26, 27 August 2018

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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