Question

By which smallest number must the following number be divided so that the quotient is a perfect cube?
 $ 7803 $

Answer 1

Hint: A quotient is the number we receive when we perform division of a number by another number. A perfect cube is any number that can be expressed as the product of only a particular number repeated three times. The easiest method to solve this problem is by factoring the given number by prime factorization method.

Complete step-by-step answer:
Let us note down what is given and what we are asked to find:
We are given a number: $ 7803 $
Our aim is to find a number that can divide the given number completely (giving no remainder), so it can be called as a factor of the given number $ 7803 $ .
But we are given a condition that the number we find, that is the factor of $ 7803 $ that we find, must divide $ 7803 $ in such a way that we get a perfect cube as its quotient. So it means that when we find the factor that gives a perfect cubed quotient, we should be able to express that quotient as a product of a specific number, multiplied three times by itself; that’s why it will be called a perfect cube. Example of a perfect cube is $ 8 $ because it can be expressed as $ 8 = 2 \times 2 \times 2 $ .
The easiest method to find such a factor would be by factoring the given number in its simplest form, using the prime factorization method. When we express the given number by the product of prime numbers, it is called the prime factorization of that number. So let us proceed to finding the factor that can give us the smallest quotient that would be a perfect cube;
Take the given number $ 7803 $ and the factorization of $ 7803 $ looks like this:
 $ \Rightarrow 7803 = 17 \times 17 \times 3 \times 3 \times 3 $
Now that we have clearly listed out the factors of $ 7803 $ , we can easily find the factors that are numbers which can be perfect cubes. For this purpose let us group the numbers which are repeated more than once;
 $ \Rightarrow 7803 = (17 \times 17) \times (3 \times 3 \times 3) $
In any perfect cube, the same number has to be multiplied three times, so it is obvious that $ 7803 $ is not a perfect cube by itself.
The only number here which is expressed in a group of three is the prime factor: $ 3 $ , giving the perfect cube as $ 27 $ . So we can ignore the condition of requiring the smallest factor that can give a perfect cubed quotient, because there is only one factor that can give a quotient which would be a perfect cube.
So we need to find a number that will divide $ 7803 $ , so that it gives us $ 27 $ . From the prime factorization, the only way to get $ 27 $ as the quotient would be to divide $ 7803 $ by:
  $ 17 \times 17 = 289 $
Once we divide $ 7803 $ by $ 289 $ , we get the perfect cube $ 27 $ as the quotient.

So hence the factor of $ 7803 $ will be $ 289 $ that gives a quotient which is a perfect cube.
So, the correct answer is “ $ 289 $ ”.

Note: Here we have used the prime factorization method so let us understand some facts about prime factorization:
- Every number, no matter how small or big, can be expressed as a product of prime numbers. In the factorization of most numbers, the same prime number can also be repeated multiple times.
- If a number cannot be factored using any prime numbers, it means that the number under consideration is itself a prime number.