3. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $1^{2}+2^{2}+4^{2}=21$ ).
Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Conversely, multiplying a divisor of $a$ by a divisor of $b$ yields a divisor of the product $a b$. This is clearly true for the squares of divisors as well (the square of a divisor of the product is equal to the product of the squares of the divisors of the factors and vice versa).

Consider the prime factorization of the number $N$: $N=p_{1}^{k_{1}} \cdot \ldots \cdot p_{n}^{k_{n}}$. Here, $p_{i}$ are distinct prime numbers, and all $k_{i} \in N$. Then $\sigma(N)=\sigma\left(p_{1}^{k_{1}}\right) \cdot \ldots \cdot \sigma\left(p_{n}^{k_{n}}\right)$ and $\sigma\left(p^{k}\right)=1+p^{2}+p^{4}+\ldots p^{2 k}$. Since $1800=2^{3} \cdot 3^{2} \cdot 5^{2}$, we have $\sigma(1800)=\left(1+2^{2}+2^{4}+2^{6}\right) \cdot\left(1+3^{2}+3^{4}\right) \cdot\left(1+5^{2}+5^{4}\right)=5035485$.

Answer: 5035485.