Question

By what smallest number we divide 2560, so the quotient becomes a perfect cube?​

Answer 1

Answer:

5

Step-by-step explanation:

$2560 = \\ 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \\ {2}^{3} \times {2}^{3} \times {2}^{3} \times 5 \\ after \: deviding by \: 5\: we \: get \: \\ {2}^{3} \times {2}^{3} \times {2}^{3} \\ = 512 \\ which \: \: is \: \: a \: \: perfect \: \: square$

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Answer 2

Answer:

Prime factorising 2560, we get,

$2560=2⁹×5$

We know, a perfect cube has multiples of 3 as powers of prime factors.

Here, number of 2's is 9 and number of 5's is 1.

So we need to multiply another

$5²$

in the factorization to make 2560 a perfect cube.

Hence, the smallest number by which 2560 must be multiplied to obtain a perfect cube is

$5²=25$

Step-by-step explanation:

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