Difference between revisions of "2012 AMC 10B Problems/Problem 10"
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== Solution 2 == | == 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 in total. | + | 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. |
~MRENTHUSIASM | ~MRENTHUSIASM |
Latest revision as of 23:08, 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
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Solution 2
Cross-multiplying gives From the prime factorization 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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