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

(New page: ==Problem== {{problem}} ==Solution== {{solution}} ==See also== {{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}})
 
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==Problem==
 
==Problem==
{{problem}}
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How many right triangles have integer leg lengths a and b and a hypotenuse of length b+1, where b<100?
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(A) 6 (B) 7 (C) 8 (D) 9 (E) 10
  
 
==Solution==
 
==Solution==
{{solution}}
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By the pytahagorean theorem, <math>a^2+b^2=b^2+2b+1</math>
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This means that <math>a^2=2b+1</math>.
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We know that <math>a,b>0</math>, and that <math>b<100</math>.
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We also know that a must be odd, since the right
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side is odd.
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So <math>a=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.
  
 
==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}}

Revision as of 17:45, 10 August 2008

Problem

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

(A) 6 (B) 7 (C) 8 (D) 9 (E) 10

Solution

By the pytahagorean 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 $\boxed{A}$.

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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