Difference between revisions of "2012 AMC 10B Problems/Problem 10"
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<math>6\times6=36</math>. | <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 | + | 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 if you reverse the order. |
<math>4\cdot 2+1=9</math> | <math>4\cdot 2+1=9</math> | ||
Revision as of 17:09, 13 July 2021
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 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
