Task 2. By which smallest natural number should the number 63000 be multiplied so that the resulting product is a perfect square?
Solution. We factorize the number 63000 into prime factors and get

$$
63000=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 7=2^{3} \cdot 3^{2} \cdot 5^{3} \cdot 7
$$

A number is a perfect square of a natural number if each number that appears in the factorization of the number into prime factors appears an even number of times, i.e., the exponent of each distinct prime factor is an even number. Since we are looking for the smallest natural number by which the given number must be multiplied, we find it as the product of the smallest factors needed to make all the exponents even numbers. Now, if we consider the right-hand side of (1), we see that we need one more factor of 2, one more factor of 5, and one more factor of 7. Thus, the required number is $2 \cdot 5 \cdot 7=70$, and we get $63000 \cdot 70=2^{4} \cdot 3^{2} \cdot 5^{4} \cdot 7^{2}=\left(2^{2} \cdot 3 \cdot 5^{2} \cdot 7\right)^{2}=2100^{2}$.