2007 AMC 10A Problems/Problem 23

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 $m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}$. Since $m \ge n > 0$, $x+y \ge x-y > 0$, from which 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 four pairs $\mathrm{(B)}$.

Solution 2

Similar to the solution above, reduce $96$ to $2^5*3^1$. To find the number of distinct factors, add $1$ to both exponents and multiply, which gives us $6*2=12$ 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 square number. Finally, subtract the two cases above for the same reason to get $\mathrm{(B)}$.

Solution 3

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 left, so the answer is $\boxed{\textbf{(B)}\; 4}$.

~HelloWorld21

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
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

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