Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.
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Answered by
9
For this first we have to find its factors and make the pair of three..
The factors are 2^9 ×5 ..
the 5 is alone and not making pair as it wants two multiples of 5 to make the pair of three so the smallest number is 5×5=25
The factors are 2^9 ×5 ..
the 5 is alone and not making pair as it wants two multiples of 5 to make the pair of three so the smallest number is 5×5=25
Answered by
15
Prime factorization of 2560 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5
By grouping in triplets, we have (2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2) * 5
As 5 is the only left out factor, we need to multiply 2560 with 2 more 5,i.e 5 * 5 = 25.
Now 2560 * 25 = 64000 which is a perfect cube.
Cube root of 64000 = (2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2) * (5 * 5 * 5) = 2 * 2 * 2 * 5 = 80
Hope it helped....
By grouping in triplets, we have (2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2) * 5
As 5 is the only left out factor, we need to multiply 2560 with 2 more 5,i.e 5 * 5 = 25.
Now 2560 * 25 = 64000 which is a perfect cube.
Cube root of 64000 = (2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2) * (5 * 5 * 5) = 2 * 2 * 2 * 5 = 80
Hope it helped....
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