1. What is the least number that can be multiplied to 69120 to make it a perfect cube?
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By: anil on 05 May 2019 02.08 am
Prime factorization of $$69120=2^9 imes3^3 imes5$$ For the number to be a perfect cube, all the exponents must be multiples of 3, thus the above number must be multiplied with $$5^2=25$$. => Ans - (C)
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