Question

what is the smallest number by which the 7803 must be multiplied so that the product are perfect cube​

Answer 1

Given,

A number 7803.

To find out,

A smallest number by which 7803 must be multiplied so that we get a perfect cube.

Solution:

The prime factorization of 7803:

$7803 = 3 \times 3 \times 3 \times 17 \times 17$

$7803 = {3}^{3} \times {17}^{2}$

Now, multiply the with the factor which is alone.

Here, 3 is in a group of three and 17 is alone.

So,

$7803 \times 17 = 132651$

Therefore 17 is a smallest number by which the 7803 is multiplied to become a perfect cube.

Verification:

Now, we have to check whether 132651 is a perfect cube or not.

The prime factorization of 132651:

$132651 = 3 \times 3 \times 3 \times 17 \times 17 \times 17$

$132651 = {3}^{3} \times {17}^{3}$

Hence it is a perfect cube.

Answer 2

Answer:

$7803$

$3 \: \: 7803 \ \\ 3 \: 2601 \\ 17 \: 867 \\ 17 \: \: \: 17 \\ 1$

prime factors of 7803

$3 \times 3 \times 3 \times 17 \times 17$

3³ multiplied 17²

ans=3³ multiplied 17² =7803

and =7803