Difference between revisions of "2007 AMC 10A Problems/Problem 17"
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==Solution 2== | ==Solution 2== | ||
− | First, we need to prime factorize <math>75</math>. <math>75</math> = <math>5^2 \cdot 3</math>. We need <math>75m</math> to be in the form <math>x^3y^3</math>. Therefore, the smallest <math>m</math> is <math>5 \cdot 3^2</math>. <math>m</math> = 45, and since <math>5^3 \cdot 3^3 = 15^3</math>, our answer is <math>45 + 15</math> = <math>\boxed (D)60</math> | + | First, we need to prime factorize <math>75</math>. <math>75</math> = <math>5^2 \cdot 3</math>. We need <math>75m</math> to be in the form <math>x^3y^3</math>. Therefore, the smallest <math>m</math> is <math>5 \cdot 3^2</math>. <math>m</math> = 45, and since <math>5^3 \cdot 3^3 = 15^3</math>, our answer is <math>45 + 15</math> = <math>\boxed {(D)60|</math> |
== See also == | == See also == |
Revision as of 15:41, 23 October 2021
Contents
Problem
Suppose that and are positive integers such that . What is the minimum possible value of ?
Solution
must be a perfect cube, so each power of a prime in the factorization for must be divisible by . Thus the minimum value of is , which makes . The minimum possible value for the sum of and is .
Solution 2
First, we need to prime factorize . = . We need to be in the form . Therefore, the smallest is . = 45, and since , our answer is = $\boxed {(D)60|$ (Error compiling LaTeX. Unknown error_msg)
See also
2007 AMC 10A (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 AMC 10 Problems and Solutions |
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