Let's find four consecutive natural numbers, where the first is divisible by 11, the second by 7, the third by 5, the fourth by 4, and furthermore, only the four digits that appear in the last place of these numbers are used in the four numbers.
I. solution. The fourth number is even, so the third is odd, thus it ends in 5, so our numbers end in $3,4,5,6$, and according to the requirement, only these four digits can appear in the higher value places as well.

Since the fourth number is divisible by 4 and its unit digit is 6, its tens digit is odd, so it is either 5 or 3. Thus, the quartet of numbers is given by one of the following two forms:
$A:$
$100 k+34$
$B: \quad 100 m+53$,
$100 k+35$,
$100 k+36$

佂

where $k$ and $m$ are suitable integers. The first form seems more practical because 33 is divisible by 11, so we look for a $k$ that is divisible by 11, whose digits are from the set $\{3,4,5,6\}$, and for which

$$
100 k+34=98 k+35+2 k-1, \quad \text { that is } \quad x=2 k-1
$$

is divisible by 7. Since $k$ is a number of the form $11 n$, then

$$
x=2 k-1=22 n-1=21 n+(n-1)
$$

is divisible by 7 if $(n-1)$ is divisible by 7. Performing the multiplication $11 n$ in the usual way, that is, bringing $k$ to the form $k=10 n+n$, we see that the first and last digit of $n$ are also from the sought set $\{3,4,5,6\}$. Thus $n$ cannot be a single-digit number because $(n-1)$ is not divisible by 7 for $n=3,4,5,6$. $n$ cannot be a two-digit number either, because in this case the middle digit of $k$ is equal to the sum of the first and last digit of $n$, and here only $3+3=6$ is possible, which would mean $n=33$, but for which $(n-1)$ is not divisible by 7.

If $n$ is a three-digit number, say

$$
n=100 a+10 b+c
$$

then $a$ and $c$ are from the set $\{3,4,5,6\}$, and to ensure that the digits of $11 n$ are also from this set, the simplest way is to choose the value of $b$ as 0. With the choice $a=3$, none of the numbers $302,303,304,305$ for $(n-1)$ are divisible by 7. Among the numbers 402, 403, 404, 405 corresponding to the value $a=4$, none are divisible by 7, but among 502, 503, 504, 505 there is: the number $(n-1)=504$. The parameter pair for this is

$$
n=505, \quad k=5555
$$

and the

$$
555533, \quad 555534, \quad 555535, \quad 555536
$$

quartet of numbers meets the requirements, except for one non-trivial divisibility: $555534: 7=79362$.

Remark. The quartet of numbers obtained is not the smallest. A smaller number in the A-type is the quartet starting with 355333.

II. solution. Now we are looking for a quartet of numbers of type B (see above). $m$ cannot be a single-digit number because none of 353, 453, 553, and 653 are divisible by 11.

$m$ cannot be a two-digit number either, because if $m=10 D+C$, then the first number can be transformed as follows:

$$
100 m+53=11(91 D+9 C+5)+(C-D-2)
$$

here the value of $C-D$ is a number between +3 and -3, so the divisibility by 11 only holds if $C-D=2$, that is, $D=C-2$. However, the number $100 m+54$ is not a multiple of 7 in either case when $C=5$ and $D=3$, or when $C=6$ and $D=4$.

However, the attempt is successful with a three-digit $m$. Indeed, if $m=100 E+10 D+C$, then the first number is:

$$
100 m+53=11(909 E+91 D+9 C+5)+(E+C-2-D)
$$

The limits of the second term are:

$$
\begin{aligned}
& E+C-2-D \geqq 3+3-2-6=-2 \text { and } \\
& E+C-2-D \leqq 6+6-2-3=7
\end{aligned}
$$

among these, only the multiple of 0 is divisible by 11, so this term is only divisible by 11 if

$$
E+C-2-D=0, \quad D=E+C-2
$$

On the other hand, in this case the second number is

$$
100 m+54=7(1428 E+143 D+14 C+8)+(4 E-D+2 C-2)
$$

here the second term, using (1), is $3 E+C$, which is a value between 12 and 24, divisible by 7 if $3 E+C=14, C=5$ and $E=3$, which leads to $D=6$ from (1). Indeed, the

$$
36553, \quad 36554, \quad 36555, \quad 36556
$$

quartet of numbers meets all the conditions. This satisfies the requirements of the problem.

Remark. It is easy to show that we get a new quartet of numbers if we write a multiple of 77 in front of the found ones, which only contains the digits 3, 4, 5, 6. Such numbers with four and five digits (from the multiplication table):

$3465, \quad 4466, \quad 4543, \quad 5544, \quad 6545, \quad$ and

$35343, \quad 36344, \quad 45353, \quad 46354, \quad 55363, \quad 56364$

and with six digits, any $F G H F G H$ number, since this is $1001 \cdot F G H=77 \cdot 13 \cdot F G H$. (Among the four- and five-digit numbers, there are also pairs with a difference of 1001 and 10010, respectively.)