Answer: 150.

Solution: The digits Й and Х are even because ЭХ and АЙ are divisible by 4. The numbers ОЙ and ОХ are less than 25, otherwise, when multiplied by 4, they would no longer be two-digit numbers. Therefore, $\mathrm{O}=2$ or $\mathrm{O}=1$. Let's consider the first case. The number 20 as $\mathrm{OX}$ or ОЙ is not suitable because 80 ends with the same digit. 22 is also not suitable due to identical digits. The only option left is 24, but the numbers ОЙ and ОХ must be different.

Thus, $\mathrm{O}=1$. The number 10 is not suitable for the same reason as 20. Taking $O X=16$, we get $А И ̆=64$, but then $A=X$, which is not allowed. For the same reason, ОЙ $=16$ is also impossible. Taking $\mathrm{OX}=14$, we find АЙ $=56$, then ОЙ $=16$, which, as we have seen, is impossible. Therefore, ОХ and ОЙ can only be 12 and 18 (in any order), and ЭХ and АЙ are then 48 and 72, respectively. We find the sum: $12+48+18+72=150$.

Comment: The answer can be obtained without finding the actual numbers. As we know, $\mathrm{O}=1$, and the digits Й and $\mathrm{X}$ are even. On the other hand, if you add a number that is four times larger to any even number, you get a number divisible by 10. Therefore, Й $+\mathrm{X}=10$. Then ОЙ $+\mathrm{OX}=10+$ $+10+10=30$, and the sum of all four numbers is five times greater, i.e., 150.
# Problem 2.

Here is a fairly simple rebus: EKH is four times greater than OI. AI is four times greater than OH. Find the sum of all four.

[4 points