6. Determine the number of ten-digit numbers in which two adjacent digits can be crossed out to obtain a number that is 99 times smaller.
SOLUTION. Let $n$ be the number satisfying the conditions of the problem. By erasing the last two digits, we reduce $n$ by at least a hundred times, so we can limit ourselves to erasing digits that are not the last. After erasing two adjacent digits, two parts of the number $n$ remain, with the first part possibly being empty if we erased its first two digits.

Let $a$ be the number determined by the first part of the number $n$ (zero in the case where the first part is empty), $b$ be the number determined by the erased two digits, and $c$ be determined by the last part of the number $n$ (let the number of digits of this part be $k$). According to the problem, we have

$$
99\left(a \cdot 10^{k}+c\right)=a \cdot 10^{k+2}+b \cdot 10^{k}+c
$$

after rearranging, $98 c=10^{k}(a+b)$. Since $c < a+b$. Additionally, the number 49 divides $a+b$, as it is itself not divisible by $10^{k}$. The positive integer quotient $(a+b) / 49$ is less than 2, so it must be equal to 1, hence $a+b=49$. From this, we derive the equality

$$
c=\frac{10^{k}}{2}=5 \cdot 10^{k-1}
$$

where the number $k$ is also determined by the number of digits of the number $a$ (if we denote the number of digits of the number $a$ by $l$, then $k=10-l-2$, and in the case of $a=0$ we naturally set $l=0$).

From the above process, it follows that for each $a \in\{0,1,2, \ldots, 49\}$ and $b=49-a$, there exists exactly one number $c$ for which the described number $n$ satisfies the conditions of the problem, and that no other such $n$ exist. We will show that all 50 such $n$ (ending in seven, six, or five zeros) are distinct.

The constructed $n$ ending in seven zeros is unique $(a=0)$. Six zeros end 9 constructed numbers $(a \in\{1,2, \ldots, 9\})$ and they are distinct, as they start with different digits. Five zeros end 40 constructed numbers $(a \in\{10,11, \ldots, 49\})$ and they are distinct, as they start with different two-digit numbers.

For clarity, let's write out a few numbers that satisfy the problem as we get them from our reasoning: for $a=0$ we have $b=49, c=50000000$ and $n=4950000000$, for $a=1$ we have $b=48, c=5000000$ and $n=1485000000$, for $a=2$ we have $n=2475000000, \ldots$, for $a=9$ we have $n=9405000000$, for $a=10$ we have $b=39, c=500000$ and $n=1039500000, \ldots$, for $a=49$ we have $b=0, c=500000$ and $n=4900500000$.

Conclusion. There are 50 numbers that satisfy the problem.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Show that by erasing the last two digits of a number with at least three digits, we get a number that is at least a hundred times smaller than the original number.

N2. A natural number is called wavy if for any three consecutive digits $a, b, c$ in its decimal representation, $(a-b)(b-c)<0$. Prove that it is possible to construct more than 25000 ten-digit wavy numbers from the digits $0,1, \ldots, 9$ that contain all the digits from zero to nine (the digit 0 cannot be in the first position). [56-B-II-3]

N3. Determine the largest two-digit number $k$ with the following property: there exists a natural number $N$, from which, after erasing the first digit from the left, we get a number that is $k$ times smaller. (After erasing the digit, the representation of the number may start with one or several zeros.) For the determined number $k$, then find the smallest suitable number $N$. [56-C-II-4]

N4. Determine the number of all triples of two-digit natural numbers $a, b, c$, whose product $a b c$ has a representation in which all digits are the same. Triples differing only in the order of the numbers are considered the same, i.e., we count them only once. [54-C-I-5]