What is the smallest positive integer K such that $2000 \times 2001 \times K$ is a perfect cube?
A.${2^3} \times {3^3} \times {23^3} \times {29^3}$
B.$2 \times 3 \times 23 \times 29$
C.$2 \times {3^2} \times {23^3} \times {29^4}$
D.${2^2} \times {3^2} \times {23^2} \times {29^2}$

Hint: In order to find the value of K, in order to make it a perfect cube, we should know what exactly a perfect cube is. A perfect cube is a number in which the integer is raised to power three, a number that can be splitted into three same integers. For example, $27$is a perfect cube as it can be splitted as $3 \times 3 \times 3$ or ${3^3}$(the integer has the power three).

Complete step-by-step answer:
We are given a number $2000 \times 2001 \times K$, we need to find the value of K, so that it makes the product a perfect cube.
For that splitting each part of the value in its factors and we get;
$2000 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$
$2001 = 3 \times 23 \times 29$
Combining them, we can write:
$2000 \times 2001 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 3 \times 23 \times 29$
Now, pairing up each integer into three same integers in each pair.
$2000 \times 2001 \times K = \left( {2 \times 2 \times 2} \right) \times 2 \times \left( {5 \times 5 \times 5} \right) \times 3 \times 23 \times 29 \times K$ …… (1)
We have one pair of three 2’s and one pair of three 5’s. And, rest we have single integers, so we need to multiply two more same integers to make it a pair of three same integers.
So, we multiply two 2’s, two 3’s, two 23’s and two 29’s in the equation 1, and consider the added value to be K. So, we get:
$K = 2 \times 2 \times 3 \times 3 \times 23 \times 23 \times 29 \times 29$
Substituting it in equation 1:
$2000 \times 2001 \times K = \left( {2 \times 2 \times 2} \right) \times 2 \times \left( {5 \times 5 \times 5} \right) \times 3 \times 23 \times 29 \times \left( {2 \times 2 \times 3 \times 3 \times 23 \times 23 \times 29 \times 29} \right)$
Now, making a pair of three integers:
$2000 \times 2001 \times K = \left( {2 \times 2 \times 2} \right) \times \left( {2 \times 2 \times 2} \right) \times \left( {5 \times 5 \times 5} \right) \times \left( {3 \times 3 \times 3} \right) \times \left( {23 \times 23 \times 23} \right) \times \left( {29 \times 29 \times 29} \right)$
We can see that now there are three same integers in each bracket, so now it’s a perfect cube formed.
So, Value of K we get as:
$K = 2 \times 2 \times 3 \times 3 \times 23 \times 23 \times 29 \times 29$
Which can be written as:
$K = {2^2} \times {3^2} \times {23^2} \times {29^2}$
Hence, Option D is correct.
So, the correct answer is “Option D”.

Note: We have taken $2 \times 2$ and others as ${2^2}$, as they have power 1, (${2^1} \times {2^1}$) and common base 2, so according to the law of radicals, we can add the powers as ${p^a}.{p^b} = {p^{a + b}}$. Similarly, we get: ${2^1} \times {2^1} = {2^{1 + 1}} = {2^2}$. Similarly, for other values we did the same.
We can check the value by multiplying the factors and then taking cube root.