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

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So <math>a=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.
 
So <math>a=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.
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I love apple pie!
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:30, 23 March 2015

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.

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

I love apple pie!

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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