Question

Find the smallest number which should be multiplied to $2916$ to get a perfect cube.

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
First of all, we just need to know such things about the perfect numbers, which is the numbers that obtain by multiplying any whole numbers (zero to infinity) twice, or the square of the given numbers yields a whole number like $\sqrt 4 = 2$ or $4 = {2^2}$ is the perfect square; similarly for the perfect cube we can define, ${3^3} = 27$ or $\sqrt[3]{{27}} = 3$ which is the perfect cubes example;

Complete step-by-step solution:
Since the given question is to find the smallest number that will surely need to multiplied to get the perfect cube as defined above is $2916$ so this number is not the perfect cube since it is not well completed and hence, we need another number to be multiplied and get the resultant;
Let $2916$ can be written as in the form of multiplication of the twos and threes like
$3 \times 3 \times 3 = 27$ Again $3 \times 3 \times 3 = 27$ and also $2 \times 2 = 4$ so if we multiplied the all terms, we get
$27 \times 27 \times 4 = 2916$ That means we can able to reform the number $2916$ as in the form of
$3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 2 \times 2 = 2916$ as we can see there are a total six three and two twos only (perfect square). Now six three can be assumed to be the perfect cube (since first three \[3\] is perfect and last three \[3\] are perfect cube) and then there are only two \[2s\] so we need another $2$ to get that number as a perfect cube; because after multiplied by two only we get three twos and then only it will be perfect cube;
Hence $2$ is the smallest number that will need to multiply by $2916$ to get that perfect square.

Note: Since if two is multiplied we get the number as $2916 \times 2 = 5832$ and then we will separate like above we did we get; $3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 2 \times 2 \times 2 = 5832$ hence all the twos are perfect square by three terms and all threes are perfect square by six terms thus the resultant.