Difference between revisions of "2007 AMC 10A Problems/Problem 23"
m (→Solution 1) |
|||
| Line 4: | Line 4: | ||
<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> | ||
| − | == Solution 1== | + | ==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> | ||
Revision as of 18:37, 9 January 2019
Contents
Problem
How many ordered pairs 
of positive integers, with 
, have the property that their squares differ by 
?
Solution 1
![\[m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3\]](http://latex.artofproblemsolving.com/b/9/d/b9dfb5d5b4fe15db2a54e11cc6df787e8cd77f3f.png)
For every two factors 
, we have 
. Since 
, 
, from which it follows that the number of ordered pairs is given by the number of ordered pairs 
. There are 
factors of , which give us six pairs 
. However, since 
are positive integers, we also need that 
are positive integers, so 
and 
must have the same parity. Thus we exclude the factors 
, and we are left with four pairs 
.
Solution 2
Similar to the solution above, reduce to 
. To find the number of distinct prime factors, add 
to both exponents and multiply, which gives us 
factors. Divide by 
since 
must be greater than or equal to 
. We don't need to worry about and being equal because is not a square number. Finally, subtract the two cases above for the same reason to get .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
