Raskina I.V.

The notorious criminal Professor Moriarty had been hiding from Sherlock Holmes and the London police for a long time. And one day, the police managed to intercept a telegram that Moriarty had sent to an accomplice:

# Meet tomorrow train STO carriage O

Inspector Lestrade had already ordered to send a police unit to look for the zero carriage of the hundredth train, but then two more intercepted telegrams to the same address were brought in:

## SECRET - OPEN = ANSWER - YOUR

SECRET - OPEN $=20010$

Lestrade was puzzled. But Holmes exclaimed:

- Now it's clear which train we need to meet!

The inspector was surprised.

- Elementary, Lestrade! - the detective explained. - This is a code. In these equations, the same letters represent the same digits, different letters represent different digits, and the dash is a minus sign! Moriarty is traveling on train number ...

Write the number of the train and the carriage.

#
Let's look at the second telegram. In the right and left parts of the equality, the last two digits are the same: ET - OY. This means that in the higher digits on the right and left, it will also be the same: SEKR - OTKR = OTB - TB.

Now, if we perform the subtractions, then on both sides, two letters will cancel out, that is, there will be two zeros at the ends. When we divide by 100, we get the following: SE - OT = O

Now let's look at the last telegram. If we write it as an addition problem in a column,

\[
\begin{aligned}
& +\frac{0 \mathrm{~T} \mathrm{~K} \mathrm{PHT}}{20010} \\
& \text{C E K P E T}
\end{aligned}
\]

we can see that OT + 2 = SE. Comparing this with the previous equality, we understand that O = 2, and then C = 3. Moreover, a carry occurred from the units place to the tens place, so T = 8 or 9.

However, if we assume that T = 8, then E = 0, and then in the addition, Y + 1 = E (i.e., Y + 1 = 0) would inevitably result in a carry to the next place. But this did not happen, so T = 9. We have learned the values of all the necessary letters.

## Answer

Train 392, carriage 2.

## Problem