Difference between revisions of "2013 UNCO Math Contest II Problems/Problem 2"
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== 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? | |
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== Solution == | == Solution == | ||
Revision as of 20:30, 15 June 2020
Problem
A number 
is equal to 
. What is the smallest positive integer 
such that the product 
is a perfect cube?
Solution
We can factor 12 into 2*2*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*12=24. Because we only have one power of three, we need two more, so we multiply 24*3*3, giving us 216, which is a perfect cube.
To find a perfect 6th power, we multiply 36*216 to get 7776. We know that the factorization of this number is 2^5*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.
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 | ||
