Difference between revisions of "2012 AMC 10B Problems/Problem 10"
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− | == Problem | + | == Problem == |
− | How many ordered pairs of positive integers (M,N) satisfy the equation <math>\frac {M}{6} | + | How many ordered pairs of positive integers <math>(M,N)</math> satisfy the equation <math>\frac{M}{6}=\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> | ||
− | + | == Solution 1 == | |
+ | Cross-multiplying gives <math>MN=36.</math> We write <math>36</math> as a product of two positive integers: | ||
+ | <cmath>\begin{align*} | ||
+ | 36 &= 1\cdot36 \\ | ||
+ | &= 2\cdot18 \\ | ||
+ | &= 3\cdot12 \\ | ||
+ | &= 4\cdot9 \\ | ||
+ | &= 6\cdot6. | ||
+ | \end{align*}</cmath> | ||
+ | The products <math>1\cdot36, 2\cdot18, 3\cdot12,</math> and <math>4\cdot9</math> each produce <math>2</math> ordered pairs <math>(M,N),</math> as we can switch the order of the factors. The product <math>6\cdot6</math> produces <math>1</math> ordered pair <math>(M,N).</math> Together, we have <math>4\cdot2+1=\boxed{\textbf{(D)}\ 9}</math> ordered pairs <math>(M,N).</math> | ||
− | + | ~Rguan (Solution) | |
− | + | ~MRENTHUSIASM (Reformatting) | |
− | + | == 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 <math>(M,N)</math> in total. | ||
− | + | ~MRENTHUSIASM | |
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==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}} |
Latest revision as of 22:08, 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
~Rguan (Solution)
~MRENTHUSIASM (Reformatting)
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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