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

(Solution)
(Solution)
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This means that <math>a^2=2b+1</math>.
 
This means that <math>a^2=2b+1</math>.
  
We know that <math>a,b>0</math>, and that <math>b<100</math>.
+
We know that <math>a,b>0</math> and that <math>b<100</math>.
  
 
We also know that a must be odd, since the right
 
We also know that a must be odd, since the right

Revision as of 13:00, 7 June 2021

Problem

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

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

Solution

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 must be odd, since the right

side is odd. An odd number (2b) added to a even number is an odd number.

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


~qkddud

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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

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