Question

What is the smallest number by which $675$ must be multiplied so that the product is a perfect cube?

Answer 1

Hint:
Start with using the method of prime factorization to express the number in form of the product of its prime factors. As we know the prime factors should have a power of multiple of $3$ for the number to be a perfect cube. Use this information to check the missing factor. Check your answer by multiplying and dividing the factors with the missing factor.

Complete Step by Step Solution:
Here in this problem, we are given a number $675$ and we need to find the smallest number that should be multiplied to get a perfect cube.
Before starting with the solution, we must understand a few concepts about a perfect cube. A perfect cube is a number that is the cube of an integer. So, the cube of every integer is called a perfect cube number. For example, ${1^3} = 1,{2^3} = 8,{3^3} = 27{\text{ and }}{4^3} = 64$ , here $1,8,27{\text{ and 64}}$ are perfect cube numbers.
For solving this problem, we can use the method of prime factorization. In this method, we use the prime factors of a number to express it in the form of the product of its prime factors along with the exponents over factors. For example, the number $56$ can be expressed as $56 = 2 \times 2 \times 2 \times 7 = {2^3} \times 7$.
For the given number $675$ we can find the prime factors step by step dividing the number by the smallest prime number possible as:

Steps	Prime Factor
Step $1$ : Dividing by $3$	$3$	\[675 \div 3 = 225\]
Step $2$ : Dividing by $3$	$3$	$225 \div 3 = 75$
Step $3$ : Dividing by $3$	$3$	$75 \div 3 = 25$
Step $4$ : Dividing by $5$	$5$	$25 \div 5 = 5$
Step $5$ : Dividing by $5$	$5$	$5 \div 5 = 1$

Therefore, the number $675$ can be written as:
$ \Rightarrow 675 = 3 \times 3 \times 3 \times 5 \times 5 = {3^3} \times {5^2}$
Now let’s multiply and divide the number with $5$as:
$ \Rightarrow 675 = {3^3} \times {5^2} = {3^3} \times {5^2} \times 5 \times \dfrac{1}{5} = {\left( {3 \times 5} \right)^3} \times \dfrac{1}{5} = {15^3} \times \dfrac{1}{5}$
Thus, we can express the given number $675$ as $675 = {15^3} \times \dfrac{1}{5}$ . So, this can easily conclude that by the multiplication of number $5$ to it, the number will become a perfect cube of $15$
$ \Rightarrow 675 \times 5 = {15^3} \times \dfrac{1}{5} \times 5 = {15^3}$

Hence, the smallest number that can be multiplied to $675$ to make it a perfect cube is $5$.

Note:
In this question, the use of the method of prime factorization was the most crucial part. An alternative approach to the question can be to find the missing number from the three repeated prime factors of the number $675$ . That is, in number $675 = \left( {3 \times 3 \times 3} \right) \times \left( {5 \times 5} \right)$ , we know that the prime factors should be in multiples of three to make this number a perfect cube.