Question

What is the least number by which $ 13720 $ must be divided so that the quotient is a perfect cube?

Answer 1

Hint: As we know that a perfect cube is a number which is equal to the number, multiplied by itself three times. We can say that if $ x $ can be a perfect cube of $ y $ , then we can write it as $ x = {y^3} $ . So in this question we will first calculate the prime factorisation of the number and then we check the number if there is any isolated term because it is the number that needs to be cleared out in order to be the perfect cube.

Complete step-by-step answer:
Here we have the number $ 13720 $ .
So the prime factorisation of
 $ 13720 = 2 \times 2 \times 2 \times 7 \times 7 \times 7 \times 5 $ .
It can also be written as $ {2^3} \times {7^3} \times 5 $ .
We can see that there is only one number which is isolated, so in order to be a perfect cube we have to divide the given number with $ 5 $ , so we can write
 $ \dfrac{{{2^3} \times {7^3} \times 5}}{5} $ .
Now it gives the value $ {2^3} \times {7^3} = 2744 $ .
Hence $ 5 $ is the least number by which $ 13720 $ must be divided so that the quotient is a perfect cube.
So, the correct answer is “5”.

Note: We should know that $ 2744 $ is the perfect cube of $ 14 $ i.e. $ 14 \times 14 \times 14 = 2744 $ . The numbers that are divisible by $ 1 $ and itself are called prime factors. We know that $ 2 $ is the smallest even prime number. We should note that in prime factorisation, we will always have prime numbers as factors. There is also another property of prime numbers which is that it must be greater than $ 1 $ .