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"

m (→‎Problem)
m (→‎Solution)
 
(12 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
 +
A number <math>x</math> is equal to <math>7\cdot24\cdot48</math>. What is the smallest positive integer <math>y</math> such that the product <math>xy</math> is a perfect cube?
  
−
EXAMPLE: The number <math>64</math> is equal to <math>8^2</math> and also equal to <math>4^3</math>, so <math>64</math> is both a perfect square and a perfect cube.
+
== Solution ==
−
 
+
We can factor <math>12</math> into <math>2 \times 2 \times 3</math>. There are already two factors of two, so we only need to multiply it by <math>3</math> to get two factors of three, giving us <math>36</math>.
−
(a) Find the smallest positive integer multiple of <math>12</math> that is a perfect square.
 
−
 
 
−
(b) Find the smallest positive integer multiple of <math>12</math> that is a perfect cube.
 
  
−
(c) Find the smallest positive integer multiple of <math>12</math> that is both a perfect square and a perfect cube.
 
  
−
== Solution ==
+
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 <math>\boxed{588}</math>.
  
 
== See Also ==
 
== See Also ==
−
{{UNC Math Contest box|n=II|year=2013|num-b=1|num-a=3}}
+
{{UNCO Math Contest box|n=II|year=2013|num-b=1|num-a=3}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Latest revision as of 18:47, 11 December 2023

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 $2 \times 2 \times 3$. 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=206848"