For an arbitrary number $x$, consider the sum $Q(x)=\lfloor x \rfloor + \left\lfloor \frac{x}{10000} \right\rfloor$. 

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Find the difference $Q(2023) - Q(2022)$. (Here $\lfloor x \rfloor$ denotes the integer part of $x$, that is, the greatest integer not exceeding $x$.)

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\$ \left\lfloor \left| \frac{\mathrm{x}-1}{\mathrm{k}} \right| \right\rfloor < \left\lfloor \left| \frac{\mathrm{x}}{\mathrm{k}} \right| \right\rfloor = m \$ means that \$ \mathrm{x}-1 < \mathrm{k} \mathrm{m} \leq \mathrm{x} \$, from which \$ x = k m \$, that is, \$ x \$

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\$\left\lfloor \frac{x}{k} \right\rfloor\$

This means that among the numbers \$ \left\lfloor 2023 \right\rfloor - \left\lfloor 2022 \right\rfloor, \left\lfloor \frac{2023}{2} \right\rfloor - \left\lfloor \frac{2022}{2} \right\rfloor, \ldots, \left\lfloor \frac{2023}{10000} \right\rfloor - \left\lfloor \frac{2022}{10000} \right\rfloor \$, there are as many ones as the number of natural divisors of 2023 that do not exceed 2023, and the rest of the numbers are zero. The difference \$ Q(2023) - Q(2022) \$ is equal to the sum of the above numbers, which is the number of natural divisors of the number \$ 2023 = 7 \cdot 17^2 \$, which is 6.

Answer

## 6.

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