Find the least number by which 4800 should be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) First,find the prime factorization of $4800$:$4800 = 48 	imes 100 = (16 	imes 3) 	imes (10^2) = (2^4 	imes 3) 	imes (2^2 	imes 5^2) = 2^6

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) First,find the prime factorization of $4800$:$4800 = 48 	imes 100 = (16 	imes 3) 	imes (10^2) = (2^4 	imes 3) 	imes (2^2 	imes 5^2) = 2^6 	imes 3^1 	imes 5^2$.To make the number a perfect cube,the exponent of each prime factor must be a multiple of $3$.In the prime factorization $2^6 	imes 3^1 	imes 5^2$:- The exponent of $2$ is $6$,which is a multiple of $3$.- The exponent of $3$ is $1$. To make it a multiple of $3$,we need to multiply by $3^2$.- The exponent of $5$ is $2$. To make it a multiple of $3$,we need to multiply by $5^1$.Therefore,the least number to be multiplied is $3^2 	imes 5^1 = 9 	imes 5 = 45$.

Find the least number by which $4800$ should be multiplied to make it a perfect cube.

Similar Questions

The least perfect cube which is completely divisible by $21, 24$ and $27$ is

$\sqrt[3]{(13.608)^{2}-(13.392)^{2}}$ is equal to

What is the value of $\sqrt[3]{\sqrt{441} + \sqrt{16} + \sqrt{4}}$?

$\sqrt[3]{4663} + 349 = ? \div 21.003$

Find the cube root of $15.625$. (in $.5$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module