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

Difference between revisions of "2007 AMC 10A Problems/Problem 23"

(→‎Solution 3)
(→‎Solution 4)
 
(One intermediate revision by one other user not shown)
Line 12: Line 12:
 
We first start as in Solution 1.
 
We first start as in Solution 1.
 
However, as an alternative, we could also "give" each of the factors a factor of <math>2.</math> This would force each one to be even. Now we have <math>\dfrac{96} {4} = 24,</math> and since <math>24= 2^3 \cdot 3,</math> the number of factors is <math>4 \cdot 2 = 8.</math> We then divide by <math>2</math> because <math>m-n \le m+n.</math> This gives <math>4,</math> as desired.
 
However, as an alternative, we could also "give" each of the factors a factor of <math>2.</math> This would force each one to be even. Now we have <math>\dfrac{96} {4} = 24,</math> and since <math>24= 2^3 \cdot 3,</math> the number of factors is <math>4 \cdot 2 = 8.</math> We then divide by <math>2</math> because <math>m-n \le m+n.</math> This gives <math>4,</math> as desired.
 +
 +
~clever14710owl
  
 
==Solution 3==
 
==Solution 3==
Line 22: Line 24:
  
 
~HelloWorld21
 
~HelloWorld21
 +
 +
 +
== Video Solution ==
 +
 +
https://www.youtube.com/watch?v=mNmXez4yvW0    ~David
  
 
== See also ==
 
== See also ==

Latest revision as of 14:24, 20 October 2024

Problem

How many ordered pairs $(m,n)$ of positive integers, with $m \ge n$, have the property that their squares differ by $96$?

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$

Solution 1

\[m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3\]

For every two factors $xy = 96$, we have $m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}$. It follows that the number of ordered pairs $(m,n)$ is given by the number of ordered pairs $(x,y): xy=96, x > y > 0$. There are $(5+1)(1+1) = 12$ factors of $96$, which give us six pairs $(x,y)$. However, since $m,n$ are positive integers, we also need that $\frac{x+y}{2}, \frac{x-y}{2}$ are positive integers, so $x$ and $y$ must have the same parity. Thus we exclude the factors $(x,y) = (1,96)(3,32)$, and we are left with $4$ pairs $\mathrm{(B)}$.

Solution 2

We first start as in Solution 1. However, as an alternative, we could also "give" each of the factors a factor of $2.$ This would force each one to be even. Now we have $\dfrac{96} {4} = 24,$ and since $24= 2^3 \cdot 3,$ the number of factors is $4 \cdot 2 = 8.$ We then divide by $2$ because $m-n \le m+n.$ This gives $4,$ as desired.

~clever14710owl

Solution 3

Similarly to the solution above, write $96$ as $2^5\cdot3^1$. To find the number of distinct factors, add $1$ to both exponents and multiply, which gives us $6\cdot2=12$ factors. Divide by $2$ since $m$ must be greater than or equal to $n$. We don't need to worry about $m$ and $n$ being equal because $96$ is not a perfect square. Finally, subtract the two cases above for the same reason to get $\mathrm{(B)}$.

Solution 4

Find all of the factor pairs of $96$: $(1,96),(2,48),(3,32),(4,24),(6,16),(8,12).$ You can eliminate $(1,96)$ and ($3,32)$ because you cannot have two numbers add to be an even number and have an odd difference at the same time without them being a decimal. You only have $4$ pairs left, so the answer is $\boxed{\textbf{(B)}\; 4}$.

~HelloWorld21


Video Solution

https://www.youtube.com/watch?v=mNmXez4yvW0 ~David

See also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_23&oldid=230187"