5. Let $d_{1}, d_{2}, \ldots, d_{n}$ be all the natural divisors of the number $10!=1 \cdot 2 \cdot \ldots \cdot 10$. Find the sum

$$
\frac{1}{d_{1}+\sqrt{10!}}+\frac{1}{d_{2}+\sqrt{10!}}+\ldots+\frac{1}{d_{n}+\sqrt{10!}}
$$
Answer: $3 /(16 \sqrt{7})$.

Let's denote the given sum as $S$. Then, since for each $d_{j}$ the number $10!/ d_{j}$ is also a divisor,

$$
S=\sum_{j=1}^{n} \frac{1}{10!/ d_{j}+\sqrt{10!}}=\frac{1}{\sqrt{10!}} \sum_{j=1}^{n} \frac{d_{j}}{\sqrt{10!}+d_{j}}
$$

Therefore, adding the original and the last expression for $S$, multiplied by $\sqrt{10!}$, we get

$$
\sqrt{10!} S+\sqrt{10!} S=\sum_{j=1}^{n}\left(\frac{\sqrt{10!}}{d_{j}+\sqrt{10!}}+\frac{d_{j}}{d_{j}+\sqrt{10!}}\right)=n
$$

Here, $n$ is the number of divisors of 10! - calculated by the formula $n=(8+1)(4+1)(2+1)(1+1)=270\left(10!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7^{1}\right.$, each prime factor can enter a divisor in any power from 0 to its power of entry in the number 10!). Thus, $S=270 /(2 \sqrt{10!})$.

Note: $d(10!)$ is the number of divisors of 10!.

$\pm$ The answer is not calculated but expressed in terms of $d(10!) .5$ points

$\pm$ The equation is set up correctly, but there is an arithmetic error in the calculation of the answer (e.g., in the calculation of $d(10!)) .5$ points

$\mp$ The solution contains the idea of setting up an equation for $S$, but due to arithmetic errors, the equation is set up incorrectly. 3 points

- The solution contains the idea of pairing terms (first with last, second with second-to-last, etc.), but further progress is absent. 2 points
- The solution computes $d(10!) .1$ point