# Problem 2.

Here is a rather simple rebus: EKH is four times greater than OY. AY is four times greater than OH. Find the sum of all four.

$[4 points]
Answer: 150.

Solution: The digits Y and X are even because EH and AI are divisible by 4. The numbers OY and OX are less than 25, otherwise, when multiplied by 4, they will no longer be two-digit numbers. Therefore, O = 2 or O = 1. Let's consider the first case. The number 20 as OX or OY is not suitable because 80 ends with the same digit. 22 is also not suitable due to identical digits. Only 24 remains, but the numbers OY and OX must be different.

Thus, O = 1. The number 10 is not suitable for the same reason as 20. Taking OX = 16, we get AI = 64, but then A = X, which should not be the case. For the same reason, OY = 16 is also impossible. Taking OX = 14, we find AI = 56, then OY = 16, which, as we have seen, is impossible. Therefore, OX and OY can only be 12 and 18 (in any order), and EH and AI 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, O = 1, and the digits Y and 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, Y + X = 10. Then OY + OX = 10 + 10 + 10 = 30, and the sum of all four numbers is five times greater, i.e., 150.