Question

By what smallest number should 3600 be multiplied so that the quotient is a perfect cube? Also, find the cube root of the quotient.

Answer 1

$\begin{array}{cc}2& 3600\\ 2& 1800\\ 2& 900\\ 2& 450\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}$
Prime factors of $3600=2\times 2\times 2\times 2\times 3\times 3\times 5\times 5$
Grouping the factors into triplets of equal factors, we get
$3600=\underset{–}{2\times 2\times 2}\times 2\times 3\times 3\times 5\times 5$
We know that, if a number is to be a perfect cube, then each of its prime factors must occur thrice.
We find that 2 occurs 4 times while 3 and 5 occurs twice only.
Hence, the smallest number, by which the given number must be multiplied in order that the product is a perfect cube $=2\times 2\times 3\times 5=60$
Also, product $=3600\times 60=216000$
Now, arranging into triplets of equal prime factors, we have
$216000=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}\times \underset{–}{3\times 3\times 3}\times \underset{–}{5\times 5\times 5}$
Taking one factor from each triplets, we get
$\sqrt[3]{3216000}=2\times 2\times 3\times 5=60$