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Difference between revisions of "2007 AMC 10A Problems/Problem 23"
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<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12</math>
 
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12</math>
  
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== Solution 1==
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==Solution 1==
 
<cmath>m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3</cmath>  
 
<cmath>m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3</cmath>  
  
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For every two factors <math>xy = 96</math>, we have <math>m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}</math>. Since <math>m \ge n > 0</math>, <math>x+y \ge x-y > 0</math>, from which it follows that the number of ordered pairs <math>(m,n)</math> is given by the number of ordered pairs <math>(x,y): xy=96, x > y > 0</math>. There are <math>(5+1)(1+1) = 12</math> factors of <math>96</math>, which give us six pairs <math>(x,y)</math>. However, since <math>m,n</math> are positive integers, we also need that <math>\frac{x+y}{2}, \frac{x-y}{2}</math> are positive integers, so <math>x</math> and <math>y</math> must have the same [[parity]]. Thus we exclude the factors <math>(x,y) = (1,96)(3,32)</math>, and we are left with four pairs <math>\mathrm{(B)}</math>.
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For every two factors <math>xy = 96</math>, we have <math>m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}</math>. It follows that the number of ordered pairs <math>(m,n)</math> is given by the number of ordered pairs <math>(x,y): xy=96, x > y > 0</math>. There are <math>(5+1)(1+1) = 12</math> factors of <math>96</math>, which give us six pairs <math>(x,y)</math>. However, since <math>m,n</math> are positive integers, we also need that <math>\frac{x+y}{2}, \frac{x-y}{2}</math> are positive integers, so <math>x</math> and <math>y</math> must have the same [[parity]]. Thus we exclude the factors <math>(x,y) = (1,96)(3,32)</math>, and we are left with <math>4</math> pairs <math>\mathrm{(B)}</math>.
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===Solution 2===
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Similar to the solution above, reduce <math>96</math> to <math>2^5*3^1</math>. To find the number of distinct prime factors, add <math>1</math> to both exponents and multiply, which gives us <math>6*2=12</math> factors. Divide by <math>2</math> since <math>m</math> must be greater than or equal to <math>n</math>. We don't need to worry about <math>m</math> and <math>n</math> being equal because <math>96</math> is not a square number. Finally, subtract the two cases above for the same reason to get <math>\mathrm{(B)}</math>.
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==Solution 2==
 +
We first start as in Solution 1.
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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.
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~clever14710owl
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==Solution 3==
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Similarly to the solution above, write <math>96</math> as <math>2^5\cdot3^1</math>. To find the number of distinct factors, add <math>1</math> to both exponents and multiply, which gives us <math>6\cdot2=12</math> factors. Divide by <math>2</math> since <math>m</math> must be greater than or equal to <math>n</math>. We don't need to worry about <math>m</math> and <math>n</math> being equal because <math>96</math> is not a perfect square. Finally, subtract the two cases above for the same reason to get <math>\mathrm{(B)}</math>.
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 +
== Solution 4 ==
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Find all of the factor pairs of <math>96</math>: <math>(1,96),(2,48),(3,32),(4,24),(6,16),(8,12).</math> You can eliminate <math>(1,96)</math> and (<math>3,32)</math> 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 <math>4</math> pairs left, so the answer is <math>\boxed{\textbf{(B)}\; 4}</math>.
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~HelloWorld21
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== Video Solution ==
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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
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All AMC 10 Problems and Solutions

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