2012 AMC 10B Problems/Problem 10: Difference between revisions
No edit summary |
|||
| Line 1: | Line 1: | ||
== 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> | ||
Revision as of 14:19, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
