Number Properties Puzzle – Challenge 20

Challenge:

If n is a perfect cube number and n = 2700m, what is one possible value of m?

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Factorize 2700:
\(2700 = 2^2 × 3^3 × 5^2\)
The product of two perfect cube numbers is another perfect cube number. For example 8 and 27 are perfect cube numbers. The product of 8 and 27 is 216, which is another perfect cube number. (8 × 27 = 216 and \(6^3 = 216\))
To make 2700m a perfect cube number we need the product of 5 and 2.
\(2700 = 2^2 × 3^3 × 5^2\)
\(2700 × 10 = 2^3 × 3^3 × 5^3\), therefore, m = 10

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