Question

find the smallest number by which 2560 must be multiplied so that the product is a perfect cube​

Answer 1

Firstly, we find prime factor.

2560=2 *2*2*2*2*2*2*2*2*5

Here,

On grouping the factor, we find that 5 is left out.

Therefor, 5 must be multiplied so that the product is a perfect cube.

I hope you like it.

Answer 2

Given,

Number = 2560

Prime Factorisation of 2560 =

$\begin{array}{r|l}2&2560\\\cline{1-2}2&1280\\\cline{1-2}2&640\\\cline{1-2}2&320\\\cline{1-2}2&160\\\cline{1-2}2&80\\\cline{1-2}2&40\\\cline{1-2}2&20\\\cline{1-2}2&10\\\cline{1-2}5&5\\\cline{1-2}&1\end{array}$

2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

In the above Prime Factorisation the number 2 is paired thrice. But the number 5 is remained single. Hence, we should multiply 5 × 5 to make perfect cube.

Thus, we should 25 to 2560 to make it a perfect cube.

2560 × 25 = 64000

3√64000 = 240.

Therefore, the perfect cube of 64000 is 240.Hence, we should multiply 25 with 2560 to make it perfect cube.