Difference between revisions of "2008 AMC 10B Problems/Problem 15"

Revision as of 10:15, 8 October 2023

How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$ ?

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$

This means that $a^2=2b+1$ .

We know that $a,b>0$ and that $b<100$ .

We also know that $a^2$ is odd and thus $a$ is odd, since the right side of the equation is odd. $2b$ is even. $2b+1$ is odd.

So $a=1,3,5,7,9,11,13$ , but if $a=1, then b=0 Thus a≠1. Thus$ (Error compiling LaTeX. Unknown error_msg)a=,3,5,7,9,11,13 $The answer is$ \boxed{A}$.

~qkddud & aopsthedude

~ pi_is_3.14

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

@@ Line 13: / Line 13: @@
 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>, but if <math>a=1, then </math>b=0 Thus a≠1.
+So <math>a=1,3,5,7,9,11,13</math>, but if <math>a=1, then b=0 Thus a≠1.
-Thus <math>a=,3,5,7,9,11,13</math>
+Thus </math>a=,3,5,7,9,11,13<math>
-The answer is <math>\boxed{A}</math>.
+The answer is </math>\boxed{A}$.