Question

Find the smallest number by which $26244$ may be divided so that the quotient is a perfect cube.

Answer 1

Hint: To find the required smallest number, we will use the prime factorization method. We will write the given number $26244$ as the multiple of primes. After that it will be written in the form of a group of three if possible. Here we need to find the smallest number such that the quotient is a perfect cube. So, we cannot make a group of two primes.

Complete step-by-step answer:
To solve the given problem, we must know the prime factorization method. By using the method of prime factorization, we can express the given number as a product of prime numbers. Therefore, we will write the given number $26244$ as the product of primes. Let us do the prime factorization of $26244$. Note that $26244$ is an even number so we can start prime factorization with prime number $2$.

$2$	$26244$
$2$	$13122$
$3$	$6561$
$3$	$2187$
$3$	$729$
$3$	$243$
$3$	$81$
$3$	$27$
$3$	$9$
$3$	$3$
	$1$

Therefore, we can write $26244 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$. Here we are dealing with a perfect cube so we have to write the obtained factorization in the form of a group of three if possible. So, we can write $26244 = 2 \times 2 \times 3 \times 3 \times \left( {3 \times 3 \times 3} \right) \times \left( {3 \times 3 \times 3} \right) \cdots \cdots \left( 1 \right)$. To find the perfect cube root, we have to take one number from each group of three but here we can see that $2 \times 2$ and $3 \times 3$ cannot be written as a group of three.
Let us divide by $2 \times 2 \times 3 \times 3 = 36$ on both sides of the equation $\left( 1 \right)$. So, we can write
$\dfrac{{26244}}{{2 \times 2 \times 3 \times 3}} = \left( {3 \times 3 \times 3} \right) \times \left( {3 \times 3 \times 3} \right)$
$ \Rightarrow \dfrac{{26244}}{{36}} = \left( {3 \times 3 \times 3} \right) \times \left( {3 \times 3 \times 3} \right)$
Here we can take one number from each group of three. So we will get a quotient which is a perfect cube. Hence, the required smallest number is $36$.

Note: Here we can say that $26244$ is a perfect square because in the prime factorization of $26244$ we can see that each prime number can be written in the form of a group of two. If the sum of all digits of a number is divisible by $3$ then that number is also divisible by $3$.