Difference between revisions of "2012 AMC 10B Problems/Problem 10"
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| − | == Problem | + | == 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> | ||
<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> | ||
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== Solution 1 == | == Solution 1 == | ||
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~MRENTHUSIASM (Reformatting) | ~MRENTHUSIASM (Reformatting) | ||
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| + | == Solution 2 == | ||
| + | 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
Contents
Problem
How many ordered pairs of positive integers 
satisfy the equation 
Solution 1
Cross-multiplying gives 
We write 
as a product of two positive integers:
The products 
and 
each produce 
ordered pairs 
as we can switch the order of the factors. The product 
produces 
ordered pair 
Together, we have 
ordered pairs
~Rguan (Solution)
~MRENTHUSIASM (Reformatting)
Solution 2
Cross-multiplying gives From the prime factorization ![\[36=2^2\cdot3^2,\]](http://latex.artofproblemsolving.com/c/1/6/c16e5afd9ff5b97f2b78e3321fe8f065575f468a.png)
we conclude that has 
positive divisors. There are 
values of 
and each value generates ordered pair So, there are 
ordered pairs 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
