## Task A-1.3.

Determine how many six-digit natural numbers exist such that by removing the first two, or the last two digits, we obtain two four-digit numbers that give the same remainder when divided by 99.
## Solution.

Let $\overline{a b c d e f}$ be a six-digit number.

The numbers $\overline{a b c d}$ and $\overline{c d e f}$ give the same remainder when divided by 99 if and only if 99 divides the number $\overline{a b c d}-\overline{c d e f}$.

Since

$$
\begin{aligned}
\overline{a b c d}-\overline{c d e f} & =100 \overline{a b}+\overline{c d}-(100 \overline{c d}+\overline{e f}) \\
& =100 \overline{a b}-99 \overline{c d}-\overline{e f} \\
& =99 \overline{a b}+\overline{a b}-99 \overline{c d}-\overline{e f} \\
& =99(\overline{a b}-\overline{c d})+\overline{a b}-\overline{e f}
\end{aligned}
$$

the number $\overline{a b c d e f}$ satisfies the given condition if and only if 99 divides $\overline{a b}-\overline{e f}$.

The numbers $\overline{a b}$ and $\overline{e f}$ are between 0 and 99, so their difference is divisible by 99 only if it is equal to 0 or 99 in absolute value.

If the difference is 0, then the numbers $\overline{a b}$ and $\overline{e f}$ are equal. In this case, $\overline{a b}$ can be chosen in 90 ways (as any number between 10 and 99).

The difference can be 99 only if $\overline{a b}=99$ and $\overline{e f}=00$.

Therefore, $\overline{a b}$ and $\overline{e f}$ can be chosen in a total of $90+1=91$ ways.

Now it remains to count the possible middle parts $\overline{c d}$ of the number $\overline{a b c d e f}$. Since $\overline{c d e f}$ is a four-digit number, $\overline{c d}$ can be any number between 10 and 99, which gives 90 ways.

Note: If a student forgets that $c$ cannot be zero when counting the possible middle parts $\overline{c d}$, and claims there are 100 possibilities for $\overline{c d}$ (from 00 to 99), in which case the final result would be $91 \cdot 100=9100$, one point should be deducted.

If a student does not consider the possibility that the difference $\overline{a b}-\overline{e f}$ can be equal to 99, in which case the final result would be $90 \cdot 90=8100$, two points should be deducted.