Difference between revisions of "2008 AMC 10B Problems/Problem 15"
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We also know that <math>a^2</math> is odd and thus <math>a</math> is odd, since the right side of the equation is odd. <math>2b</math> is even. <math>2b+1</math> is odd. | We also know that <math>a^2</math> is odd and thus <math>a</math> is odd, since the right side of the equation is odd. <math>2b</math> is even. <math>2b+1</math> is odd. | ||
− | So <math>a=1,3,5,7,9,11,13</math>, | + | So <math>a=1,3,5,7,9,11,13</math>, but if <math>a=1, then </math>b=0 Thus a≠1. |
− | Thus | + | Thus <math>a=,3,5,7,9,11,13</math> |
+ | The answer is <math>\boxed{A}</math>. | ||
− | ~qkddud & aopsthedude | + | ~qkddud & aopsthedude |
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 11:15, 8 October 2023
Problem
How many right triangles have integer leg lengths and
and a hypotenuse of length
, where
?
Solution
By the Pythagorean theorem,
This means that .
We know that and that
.
We also know that is odd and thus
is odd, since the right side of the equation is odd.
is even.
is odd.
So , but if
b=0 Thus a≠1.
Thus
The answer is
.
~qkddud & aopsthedude
Video Solution by OmegaLearn
https://youtu.be/euz1azVKUYs?t=135
~ pi_is_3.14
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.