find the smallest number by which 2560 must be multiplied so that the product is a perfect cube
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prime factors of 2560 =2^9× 5
prime factors of 2560= (2^3)^3 ×5
2 is in the power of 3 but 5 is not so we multiply by 5^2(25) to make it a perfect cube
the smallest number is 25
prime factors of 2560= (2^3)^3 ×5
2 is in the power of 3 but 5 is not so we multiply by 5^2(25) to make it a perfect cube
the smallest number is 25
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