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2007 AMC 10A Problems/Problem 23

Revision as of 21:01, 25 May 2018 by Zeric (talk | contribs) (→‎Solution 1: -Add solution)

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 prime 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

First, write $m$ as $n+x$. \[m^2-n^2=96 \Longrightarrow (n+x)^2-n^2=96 \Rightarrow 2nx+x^2=96\] We see that $2nx$ and $96$ have the same parity, therefore $x^2$ is even. $0<x<10$ because $0<x^2<96$. The possible values of $x$ are ${2,4,6,8}$. For every value of $x$, we can find the corresponding integer value of $n$ when $n>0$ Therefore, there are 4 possible values of n.


When $x=2$, $n=23$, when $x=4$, $n=10$, when $x=6$, $n=5$, when $x=8$, $n=2$

~ZericHang

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

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