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

Revision as of 12:03, 7 June 2021

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 $a=3,5,7,9,11,13$ , and the answer is $\boxed{A}$ .

~qkddud

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

@@ Line 11: / Line 11: @@
 We know that <math>a,b>0</math> and that <math>b<100</math>.
-We also know that <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=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.