2008 AMC 10B Problems/Problem 15: Difference between revisions

Revision as of 16:45, 10 August 2008

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

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 $\boxed{A}$ .

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 1: / Line 1: @@
 ==Problem==
-{{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==
-{{solution}}
+By the pytahagorean theorem, <math>a^2+b^2=b^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 also know that a must be odd, since the right
+side is odd.
+So <math>a=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.
 ==See also==
 {{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}}