To determine the values of the digits $x, y, z, u, v$ such that the number

$x 61 y 064 z u v$

written in the decimal system is divisible by 61875.
I. solution: Factorize the divisor into prime factors:

$$
61875=3^{2} \cdot 5^{4} \cdot 11=9 \cdot 11 \cdot 625
$$

Since these factors are pairwise relatively prime, it is necessary and sufficient for the number we are looking for to be divisible by each of these factors separately.

Numbers are divisible by 625 if and only if the number formed by their last four digits is also divisible by 625. (See "K. M. L." March 1953 issue, page 88, problem 70.) Between 4000 and 4999, 625 has only one multiple:

$$
7 \cdot 625=4375, \quad \text { so } \quad 4 z u v=4375 \text {. }
$$

When divided by 9, every number gives the same remainder as the sum of its digits, so it is necessary that

$$
x+6+1+y+0+6+4+3+7+5=x+y+32=9 m
$$

When divided by 11, every number gives the same remainder as the number obtained by adding every second digit starting from the units place and subtracting the sum of the omitted digits (See "K. M. L." March 1953 issue, page 81, problem 475.) In this case,

$$
\begin{gathered}
(5+3+6+y+6)-(7+4+0+1+x)=(20+y)-(12+x)= \\
=8+y-x=11 n
\end{gathered}
$$

Thus,

$$
x+y=9 m-32 \quad \text { and } \quad-x+y=11 n-8
$$

Here, $x$ and $y$ are single-digit numbers, so

$$
0 \leq x+y \leq 18, \quad-9 \leq-x+y \leq 9
$$

and thus $m=4$ or $5, n=0$ or 1. Since the sum and difference of two integers cannot be one odd and the other even, $m=4, n=0$ or $m=5, n=1$ can satisfy. In the first case, a negative number would result for $y$, so only $x+y=13$ and $-x+y=3$ are possible, from which $x=5$ and $y=8$.

Thus, the number we are looking for is

$$
5618064375
$$

and it must indeed be divisible by 61875, because the choice of the last three digits makes it divisible by 625, and the choice of $x$ and $y$ makes it divisible by 9 and 11.

II. solution: If the number is divisible by 61875, then, for example, its 16-fold is divisible by $16 \cdot 61875=990000$, so it ends in four 0s in all cases. The last four digits of the number in question, due to $4 \cdot 16=64$, are the same as those of

$$
4000+16 \cdot z u v
$$

Since $16 \cdot z u v<16000$, we can only get a number ending in four 0s if

$$
16 \cdot z u v=6000, \quad \text { or } \quad z u v=375
$$

In addition to this, the 16-fold of the number in question must also be divisible by 99. Since 16 and 99 are relatively prime, this only follows if the original number is also divisible by 99. There is a simple divisibility rule for 99. (See "K. M. L." April 1953 issue, page 118, problem 74.) Since

$$
\begin{gathered}
100=99+1, \quad 10000=99 \cdot 101+1,1000000=99 \cdot 10101+1 \\
100000000=99 \cdot 1010101+1, \ldots
\end{gathered}
$$

the number in question gives the same remainder when divided by 99 as the sum of the following two-digit numbers:

$$
x 6+1 y+06+43+75=x y+140
$$

This number is greater than 140 and less than 240, so it can only be divisible by 99 if it equals $2 \cdot 99=198$, in which case

$$
x y=58, \quad \text { so } \quad x=5, y=8
$$

and thus the number we are looking for is

$$
5618064375
$$

and this is indeed divisible by 61875, because it is divisible by 99, so its 16-fold is also divisible by it, and the 16-fold of this number also ends in four 0s. Thus, the 16-fold of the number is divisible by 990000, and therefore the original number is divisible by 61875.