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Difference between revisions of "2013 UNCO Math Contest II Problems/Problem 2"

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== Solution ==
 
== Solution ==
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We can factor <math>12</math> into <math>2 \times 2 \times 2</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>.
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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>.
  
  

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
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