7. Kopyatych took some natural number, raised it to the 1st, 2nd, 3rd, 4th, and 5th powers. Then he encrypted the numbers by replacing the same digits with the same letters and different digits with different letters. He wrote each encrypted number on a separate piece of paper. But Nyusha left only a fragment of each piece of paper, preserving only the last two letters, and even those were mixed up. The result was: OK, SY, YZ, SH, LI. Find the value of the expression

$$
\text { Yo } \cdot Z \cdot I \cdot \mathrm{K}-L \cdot O \cdot \mathrm{C} \cdot \text { Y. Sh. }
$$
# Solution:

It is clear that the letters К, Ё, Ж, И are of the same parity, and among them there are no 0 or 5.

If К, Ё, Ж, И are even, then all powers except the first one must be divisible by 4, and then О, Ш are also even digits, which is not good.

The last digit of the number in the first and fifth power is the same, so ШИ are the last two digits of the first and fifth power of the number. But then the fourth power of this number ends in 01.

Then, up to the permutation of the factors,

$$
\text { Ё } \cdot \text { K } \cdot \text { И } \cdot \mathrm{K}-Л \cdot \mathrm{O} \cdot \mathrm{C} \cdot \text { Я } \cdot \amalg=1 \cdot 3 \cdot 7 \cdot 9-0 \cdot 2 \cdot 4 \cdot 6 \cdot 8=189 .
$$

Remark. There are examples in this problem, and not just one. There are $07,49,43,01,07$ and $43,49,07,01$, 43. It is important that we cannot determine which letter is Ш and which is О.