Difference between revisions of "2009 AMC 8 Problems/Problem 17"

Revision as of 22:02, 17 January 2025

Problem

The positive integers $x$ and $y$ are the two smallest positive integers for which the product of $360$ and $x$ is a square and the product of $360$ and $y$ is a cube. What is the sum of $x$ and $y$ ?

$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 85 \qquad \textbf{(C)}\ 115 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 610$

Video Solution (XXX)

https://www.youtube.com/watch?v=ZuSJdf1zWYw ~David

:)

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 ==Video Solution (XXX)==
 https://www.youtube.com/watch?v=ZuSJdf1zWYw   ~David
-==Solution==
-The prime factorization of <math>360=2^3 \cdot 3^2 \cdot 5</math>. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is the minimum possible value of x. Similarly, y can be found by making all the exponents divisible by 3, so the minimum possible value of <math>y</math> is <math>3 \cdot 5^2=75</math>. Thus, our answer is <math>x+y=10+75=\boxed{\textbf{(B)}\ 85}</math>.
 ==:)==

Page

Toolbox

Search

Difference between revisions of "2009 AMC 8 Problems/Problem 17"

Revision as of 22:02, 17 January 2025

Contents

Problem

Video Solution (XXX)

:)

See Also