Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 8 Problems/Problem 17"
(→‎Solution)
(4 intermediate revisions by 4 users not shown)
Line 8: Line 8:
 
\textbf{(D)}\    165  \qquad
 
\textbf{(D)}\    165  \qquad
 
\textbf{(E)}\    610</math>
 
\textbf{(E)}\    610</math>
 +
 +
==Video Solution==
 +
https://youtu.be/7an5wU9Q5hk?t=2768
  
 
==Solution==
 
==Solution==
−
The prime factorization of <math>360=2^3*3^2*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 x. Similarly, y can be found by making all the exponents divisible by 3, so <math>y=3*5^2=75</math>. Thus x+y=85, B.
+
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>.
 +
 
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=16|num-a=18}}
 
{{AMC8 box|year=2009|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Revision as of 20:22, 27 October 2020

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

https://youtu.be/7an5wU9Q5hk?t=2768

Solution

The prime factorization of $360=2^3 \cdot 3^2 \cdot 5$. 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 $y$ is $3 \cdot 5^2=75$. Thus, our answer is $x+y=10+75=\boxed{\textbf{(B)}\ 85}$.

See Also

2009 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_17&oldid=135955"