Difference between revisions of "2007 AMC 10A Problems/Problem 23"
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− | Similarly to the solution above, write <math>96</math> as <math>2^5\cdot3^1</math>. To find the number of distinct factors, add <math>1</math> to both exponents and multiply, which gives us <math>6 | + | Similarly to the solution above, write <math>96</math> as <math>2^5\cdot3^1</math>. To find the number of distinct factors, add <math>1</math> to both exponents and multiply, which gives us <math>6\cdot2=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 perfect square. Finally, subtract the two cases above for the same reason to get <math>\mathrm{(B)}</math>. |
== Solution 3 == | == Solution 3 == |
Revision as of 12:38, 12 April 2022
Contents
[hide]Problem
How many ordered pairs of positive integers, with , have the property that their squares differ by ?
Solution 1
For every two factors , we have . 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 pairs .
Solution 2
Similarly to the solution above, write as . To find the number of distinct 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 perfect square. Finally, subtract the two cases above for the same reason to get .
Solution 3
Find all of the factor pairs of : You can eliminate and ( 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 pairs left, so the answer is .
~HelloWorld21
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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