2012 AMC 10B Problems/Problem 10: Difference between revisions
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[[Category: Introductory Number Theory Problems]] | |||
Latest revision as of 17:54, 19 October 2025
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 | ||
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