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 "2013 UNCO Math Contest II Problems/Problem 2"

(→‎Solution)
(→‎Solution)
Line 9: Line 9:
 
To find the perfect cube, we need all of the prime factors to be to the third power. Because <math>2</math> is squared, we need to multiply by a power of <math>2</math>, giving us <math>2 \times 12</math>, which is <math>24</math>. Because we only have one power of three, we need two more, so we multiply <math>24 \times 3 \times 3</math>, giving us <math>216</math>, which is a perfect cube.
 
To find the perfect cube, we need all of the prime factors to be to the third power. Because <math>2</math> is squared, we need to multiply by a power of <math>2</math>, giving us <math>2 \times 12</math>, which is <math>24</math>. Because we only have one power of three, we need two more, so we multiply <math>24 \times 3 \times 3</math>, giving us <math>216</math>, which is a perfect cube.
  
−
To find a perfect 6th power, we multiply <math>36</math> by <math>216</math> to get <math>7776</math>. We know that the factorization of this number is <math>2^5 \times 3^5</math>. This means that this number is <math>6^5</math>. We need a perfect 6th, so we multiply by <math>6</math> to get <math>46656</math>, which is <math>6^6</math>, or \boxed {588}.
+
To find a perfect 6th power, we multiply <math>36</math> by <math>216</math> to get <math>7776</math>. We know that the factorization of this number is <math>2^5 \times 3^5</math>. This means that this number is <math>6^5</math>. We need a perfect 6th, so we multiply by <math>6</math> to get <math>46656</math>, which is <math>6^6</math>, or <math>\boxed{588}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 00:06, 15 February 2021

Problem

A number $x$ is equal to $7\cdot24\cdot48$. What is the smallest positive integer $y$ such that the product $xy$ is a perfect cube?

Solution

We can factor $12$ into $2 \times 2 \times 2$. There are already two factors of two, so we only need to multiply it by $3$ to get two factors of three, giving us $36$.


To find the perfect cube, we need all of the prime factors to be to the third power. Because $2$ is squared, we need to multiply by a power of $2$, giving us $2 \times 12$, which is $24$. Because we only have one power of three, we need two more, so we multiply $24 \times 3 \times 3$, giving us $216$, which is a perfect cube.

To find a perfect 6th power, we multiply $36$ by $216$ to get $7776$. We know that the factorization of this number is $2^5 \times 3^5$. This means that this number is $6^5$. We need a perfect 6th, so we multiply by $6$ to get $46656$, which is $6^6$, or $\boxed{588}$.

See Also

2013 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_UNCO_Math_Contest_II_Problems/Problem_2&oldid=146741"