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> | + | 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> | ||
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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 | + | <math>4\cdot 2+1=9</math> |
<math>\boxed{\textbf{(D)}\ 9}</math> | <math>\boxed{\textbf{(D)}\ 9}</math> | ||
+ | |||
==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 15:19, 28 August 2020
Problem 10
How many ordered pairs of positive integers satisfy the equation = ?
Solution
=
is a ratio; therefore, you can cross-multiply.
Now you find all the factors of 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.
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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