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

Revision as of 21:01, 25 May 2018

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

@@ Line 10: / Line 10: @@
 ===Solution 2===
 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>.
+===Solution 3===
+First, write <math>m</math> as <math>n+x</math>. <cmath>m^2-n^2=96 \Longrightarrow (n+x)^2-n^2=96 \Rightarrow 2nx+x^2=96</cmath>
+We see that <math>2nx</math> and <math>96</math> have the same parity, therefore <math>x^2</math> is even. <math>0<x<10</math> because <math>0<x^2<96</math>. The possible values of <math>x</math> are <math>{2,4,6,8}</math>. For every value of <math>x</math>, we can find the corresponding integer value of <math>n</math> when <math>n>0</math> Therefore, there are 4 possible values of n.
+When <math>x=2</math>, <math>n=23</math>, when <math>x=4</math>, <math>n=10</math>, when <math>x=6</math>, <math>n=5</math>, when <math>x=8</math>, <math>n=2</math>
+~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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