Task 5. Replace in the equation

$$
\text { PIE = SLICE + SLICE + SLICE + ... + SLICE }
$$

identical letters with identical digits, and different letters with different digits, so that the equation is true, and the number of "slices of pie" is the largest possible.
Answer. The maximum number of "pieces" is seven, for example: ПИРОГ $=95207$, КУСОК $=13601$.

Solution. An example for seven "pieces" is given above. We will show that there cannot be more than seven "pieces". For this, it is convenient to rewrite the condition as a multiplication example: ПИРОГ $=$ КУСОК $\cdot n$, where $n$ is the number of "pieces".

It is clear that if there are 10 or more "pieces", the right side will exceed 100000, so there will be no solutions.

A solution is also impossible for nine "pieces". For ПИРОГ $=$ КУСОК $\cdot 9$ to be a five-digit number, $К$ must be 1. But then ПИРОГ starts and ends with a nine, which should not be the case.

Now we will prove that there cannot be eight "pieces". Let КУСОК $\cdot 8=$ ПИРОГ. Since all the digits in ПИРОГ are different, ПИРОГ $\leqslant 98765$, and then КУСОК $\leqslant 98765 / 8=12345.675$, so КУСОК $\leqslant 12345$. Then it is clear that $К=1$ and $\Gamma=8$. The letter О represents a digit, the product of which by 8 ends in the same digit. It is easy to see that this can only be zero. Since the digits 0 and 1 are already used, and $\mathrm{V} \leqslant 2$ and $\mathrm{C} \leqslant 3$, then $\mathrm{V}=2$ and $\mathrm{C}=3$. Thus, there is only one possibility: КУСОК $=12301$. But this number does not work, as $12301 \cdot 8=98408$, and in 98408 the digits are repeated, which should not be the case.

The example given for seven "pieces" is not the only one; there are three more: $14051 \cdot 7=98357, 12351 \cdot 7=86457, 12051 \cdot 7=84357$. Of course, it was sufficient to provide one example.