Question

By what least number should 153 be divided to get a perfect cube.

Answer 1

Hint: According to the given question, we will start making factors of 1536. That means we will start with the smallest prime number of 1536 till the quotient becomes 1. Hence, take all the factors from which number you are dividing and make all the triplet pairs to get the cube root and we will divide the number which is left and does not make a triplet.

Complete step-by-step answer:
As, we have to calculate the cube root of 1536 by factorization method.
Here, we will start dividing the number with the smallest prime number that is 2 till it leaves quotient 1.
So, we will start with the given value that is 1536.
$\Rightarrow$ 1536 = \[\dfrac{{1536}}{2}\]
On dividing with 2 we get = 768
Here, again we will continue dividing the quotient that is 768.
$\Rightarrow$ 768 = \[\dfrac{{768}}{2}\]
On dividing with 2 we get = 384
Here, again we will continue dividing the quotient that is 384.
$\Rightarrow$ 384 = \[\dfrac{{384}}{2}\]
On dividing with 2 we get = 192
Here, again we will continue dividing the quotient that is 192.
$\Rightarrow$ 192 = \[\dfrac{{192}}{2}\]
On dividing with 2 we get = 96
Here, again we will continue dividing the quotient that is 96.
$\Rightarrow$ 96 = \[\dfrac{{96}}{2}\]
On dividing with 2 we get = 48
Here, again we will continue dividing the quotient that is 48.
$\Rightarrow$ 48 = \[\dfrac{{48}}{2}\]
On dividing with 2 we get = 24
Here, again we will continue dividing the quotient that is 24.
$\Rightarrow$ 24 = \[\dfrac{{24}}{2}\]
On dividing with 2 we get = 12
Here, again we will continue dividing the quotient that is 12.
$\Rightarrow$ 12 = \[\dfrac{{12}}{2}\]
On dividing with 2 we get = 6
Here, again we will continue dividing the quotient that is 6.
$\Rightarrow$ 6 = \[\dfrac{6}{2}\]
On dividing with 2 we get = 3.
Now, as 3 is not divisible by 2. So we move on to next prime number that is 3
$\Rightarrow$ 3 = \[\dfrac{3}{3}\]
On dividing with 3 we get = 1 which is our required quotient.
Therefore, Calculated Prime Factors of 1536 are
$\Rightarrow$ \[1536 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3\;\]
On simplifying the factors we get
$\Rightarrow$ \[1536 = {2^9} \times {3^1}\]
As, 3 has been left which is not having a triplet so we will divide 1536 by 3 to get a perfect cube that is \[\dfrac{{1536}}{3} = 512\]
Hence, \[\sqrt[3]{{512}} = 8\] which is a perfect cube.

Note: To solve these types of questions, we must remember to find all the factors of the given number in the prime number. Similarly we can find out the square root of the given number by just taking out pairs of the prime number from the calculated factors. Hence, to find a perfect square or perfect cube just divide with a number which does not make a pair or triplet respectively.