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

Latest revision as of 10:39, 8 October 2023

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 $a=1,3,5,7,9,11,13$ , but if $a=1$ , then $b=0$ . Thus $a\neq1.$

$a=3,5,7,9,11,13$

The answer is $\boxed{A}$ .

~qkddud (edited by aopsthedude and bburubburu)

~ pi_is_3.14

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 13: / Line 13: @@
 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>.
+So <math>a=1,3,5,7,9,11,13</math>, but if <math>a=1</math>, then <math>b=0</math>. Thus <math>a\neq1.</math>
+<math>a=3,5,7,9,11,13</math>
+The answer is <math>\boxed{A}</math>.
-~qkddud
+~qkddud (edited by aopsthedude and bburubburu)
+== Video Solution by OmegaLearn ==
+https://youtu.be/euz1azVKUYs?t=135
+~ pi_is_3.14
 ==See also==
 {{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}}
 {{MAA Notice}}