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

(Created page with "== Problem 10 == How many ordered pairs of positive integers (M,N) satisfy the equation <math>\frac {M}{6}</math> = <math>\frac{6}{N}</math> <math> \textbf{(A)}\ 6\qquad...")
 
(Solution)
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Now you find all the factors of 36:
 
Now you find all the factors of 36:
  
1*36=36
+
<math>1\times36=36</math>
  
2*18=36
+
<math>2\times18=36</math>
  
3*12=36
+
<math>3\times12=36</math>
  
4*9=36
+
<math>4\times9=36</math>
  
6*6=36.
+
<math>6\times6=36</math>.
  
 
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.

Revision as of 17:54, 24 February 2012

Problem 10

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

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\10$ (Error compiling LaTeX. ! Undefined control sequence.)

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.

$4*2+1=\boxed{9}$

OR

$\textbf{(D)}$

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