5. Determine the largest number obtained by deleting 100 digits from the number 12345678910111213...99100, whose digits are the natural numbers from 1 to 100 arranged from left to right.
Solution. There are 0 one-digit numbers. From 10 to 99, there are 90 two-digit numbers. Thus, the given number has $9 + 2 \cdot 90 + 3 = 192$ digits. After deleting 100 digits, we need to obtain a 92-digit number. For any two numbers with the same number of digits, the number with the larger first digit is greater. Clearly, the number we are looking for should start with as many 9s as possible. Therefore, we first delete the first 8 digits. Then we delete the sequence of digits $101112 \ldots 181$, which consists of $9 \cdot 2 + 1 = 19$ digits. Similarly, we delete the sequences 202122...282, 303132...383, 404142...484. Thus, $8 + 19 \cdot 4 = 84$ digits have been deleted, and the number obtained is 999950515253...99100. We need to delete 16 more digits. If we delete the sequence 505152...57, which consists of 16 digits, we get the number 9999958596061...99100. However, the largest number is obtained by deleting the sequence 505152...565, which consists of 15 digits, so that the next digit we delete is not 7 but 5 from 58. Thus, the desired number is 9999978596061...99100.

## VIII Section - 8th Grade