Zsspasii A.A.

The number of the current Olympiad (70) is formed by the last digits of the year of its holding, written in reverse order.

How many more times will such a situation occur in this millennium?

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Let in some year the described coincidence occurred.

If the number of the Olympics is two-digit, then the sum of this number and the number formed by the last two digits of the year is divisible by 11 (the sum of two numbers consisting of digits $a$ and $b$ is $11(a+b)$). Since each year this sum increases by 2, the event can repeat no more often than every 11 years. Indeed, the 81st and 92nd Olympics will take place in 2018 and 2029.

If the number of the Olympics is three-digit, then the second-to-last digits of the number and the year coincide. Therefore, the second-to-last digit of their difference can only be 0 or 9. But the difference between the year of the Olympics and its number is always 1937. Contradiction.

If the number of the Olympics is four-digit, then the sums of the digits of the number and the year coincide. Since any number gives the same remainder when divided by 9 as the sum of its digits, the difference between the year and the number must be divisible by 9. But the remainder when 1937 is divided by 9 is 2. Contradiction.

Note that such a situation will not be observed in the future.

## Answer

Twice.