Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2012 AMC 10B Problems/Problem 10: Difference between revisions

← Older editNewer edit →
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
No edit summary
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
Line 20: Line 20:


== 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

Revision as of 22:08, 3 September 2021

Problem

How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution 1

Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers: \begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*} The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the factors. The product $6\cdot6$ produces $1$ ordered pair $(M,N).$ Together, we have $4\cdot2+1=\boxed{\textbf{(D)}\ 9}$ ordered pairs $(M,N).$

~Rguan (Solution)

~MRENTHUSIASM (Reformatting)

Solution 2

Cross-multiplying gives $MN=36.$ From the prime factorization \[36=2^2\cdot3^2,\] we conclude that $36$ has $(2+1)(2+1)=9$ positive divisors. There are $9$ values of $M,$ and each value generates $1$ ordered pair $(M,N).$ So, there are $\boxed{\textbf{(D)}\ 9}$ ordered pairs $(M,N)$ 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_10&oldid=161527"