Difference between revisions of "2007 AMC 10A Problems/Problem 17"
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− | <math>3 \cdot 5^2m</math> must be a perfect cube, so each power of a prime in the factorization for <math>3 \cdot 5^2m</math> must be divisible by <math>3</math>. Thus the minimum value of <math>m</math> is <math>3^2 \cdot 5 = 45</math>, which makes <math>n = \sqrt[3]{3^3 \cdot 5^3} = 15</math>. | + | <math>3 \cdot 5^2m</math> must be a perfect cube, so each power of a prime in the factorization for <math>3 \cdot 5^2m</math> must be divisible by <math>3</math>. Thus the minimum value of <math>m</math> is <math>3^2 \cdot 5 = 45</math>, which makes <math>n = \sqrt[3]{3^3 \cdot 5^3} = 15</math>. The minimum possible value for the sum of <math>m</math> and <math>n</math> is <math>60\ \mathrm{(D)}</math>. |
== See also == | == See also == |
Revision as of 14:40, 3 June 2021
Problem
Suppose that and
are positive integers such that
. What is a 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
.
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
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