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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 ==
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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>  
  
−
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>.  
+
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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===Solution 2===
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Similar to the solution above, reduce <math>96</math> into <math>2^5*3^1</math>. To find the number of disinct 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>.
  
 
== See also ==
 
== See also ==

Revision as of 03:44, 19 June 2017

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 $96$ into $2^5*3^1$. To find the number of disinct 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 $\mathrm{(B)}$.

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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