Egy, a tízes számrendszerben felírt valódi hatjegyú számról a következőket tudjuk: 
a) egyik jegye 7, 
b) osztható 9-cel, 
c) ha sorra kivonjuk belőle az $i$-edik és $j$-edik jegyének felcserélésével keletkező számokat, a nullától különböző különbségek között van 2525-tel, 2168-cal, 4375-tel és 6875 -tel osztható. 
Melyik ez a szám?

A real six-digit number, written in the decimal system, is known to have the following properties:
a) one of its digits is 7,
b) it is divisible by 9,
c) if we sequentially subtract from it the numbers formed by swapping the $i$-th and $j$-th digits, the non-zero differences include numbers divisible by 2525, 2168, 4375, and 6875.
Which is this number?
Let the sought number be $\overline{a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}}$. If we subtract the number obtained by swapping the $i$-th and $j$-th digits $(i>j)$ from the original number, the difference is

$$
10^{i} a_{i}+10^{j} a_{j}-10^{j} a_{i}-10^{i} a_{j}=\left(10^{i}-10^{j}\right)\left(a_{i}-a_{j}\right)=2^{j} \cdot 5^{j} \cdot 9 \cdot(\underbrace{11 \ldots 1}_{i-j \text { digits }})\left(a_{i}-a_{j}\right)
$$

The number consisting of $i-j$ ones can be the following:

$$
i=j+11 i=j+211 i=j+3111=3 \cdot 37 i=j+41111=11 \cdot 101 i=j+511111=41 \cdot 271
$$

Let's examine how the desired properties of the differences can arise! First, consider the number divisible by $2168=2^{3} \cdot 271$. We use the (1) decomposition. Since we are dealing with a real difference and $\left|a_{i}-a_{j}\right|<10$, the difference can only be divisible by 271 if $i=j+5$, i.e., $i=5, j=0$. In this case, however, $8 \mid a_{i}-a_{j}$, which is possible in the following cases $\left(a_{0} \neq a_{5}\right.$, because we are considering non-zero differences, and $a_{5} \neq 0$, because we are looking for a real six-digit number):

$$
\overline{8 a_{4} a_{3} a_{2} a_{1} 0}, \quad \overline{9 a_{4} a_{3} a_{2} a_{1} 1}, \quad \overline{1 a_{4} a_{3} a_{2} a_{1} 9}
$$

Now let's consider $2525=5^{2} \cdot 101$: divisibility by 101 can only come from 1111, so according to (2), $i=j+4$. For $i=4, j=0$, $25 \mid a_{i}-a_{j}$ would need to hold, which is impossible, so $i=5, j=1$ and $5 \mid a_{1}-a_{5}$. The possibilities are:

$$
\overline{8 a_{4} a_{3} a_{2} 30}, \quad \overline{9 a_{4} a_{3} a_{2} 41}, \quad \overline{1 a_{4} a_{3} a_{2} 69}
$$

For $6875=5^{4} \cdot 11$, only $j \geq 3$ is possible, since $a_{i}-a_{j}$ can be divisible by at most the first power of 5. For divisibility by 11, $i-j=2$ or 4, which together with $j \geq 3$ leads to the case $i=5, j=3$. In this case, $5 \mid a_{3}-a_{5}$ is also necessary, so the numbers are:

$$
\overline{8 a_{4} 3 a_{2} 30}, \quad \overline{9 a_{4} 4 a_{2} 41}, \quad \overline{1 a_{4} 6 a_{2} 69}
$$

When examining $4375=7 \cdot 5^{4}$, similarly to the previous case, $j \geq 3$ is obtained. Since 7 can only divide $a_{i}-a_{j}$, $j=3$ can be ruled out: in this case, $35 \mid a_{i}-a_{j}$ would need to hold.

The remaining case is $j=4, i=5$, and $7 \mid a_{4}-a_{5}$:

$$
\overline{813 a_{2} 30}, \quad \overline{924 a_{2} 41}, \quad \overline{186 a_{2} 69}
$$

We also know that the number has a 7 as one of its digits, which in all three cases can only be $a_{2}$. The divisibility by 9 is satisfied only by the number 924741, so this is the only possible number. We have already seen that this number indeed meets the criteria.

Based on the work of Gyula Pap (Debrecen, Fazekas M. Gymnasium, II. grade)