From the 1954 Moscow Maths Olympiad:

*Delete 100 digits from the number 123456789101112…979899100 (that is, the number formed by writing the first 100 positive integers next to each other).*

*What is the largest possible number that could remain?*

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This is one of those rare problems that I just had to try before doing anything else! The premise is so simple, and the prerequisite knowledge so minimal, that you could give this problem to someone aged 9 or 99 and they could make a decent attempt at it.

The original number contains 192 digits, so deleting 100 of them will leave a 92-digit number.

Idea: get as many digit 9s to the front as I possibly can.

The first five digit 9s in the original number come from 9, 19, 29, 39, 49. I could pick these five 9s to be the first five digits of my new number. That would leave the 103 digits “50515253…9899100” that come after the 49, from which I need to choose 87 digits (to make the total number of digits 92). So my new number will definitely start “99999…”

Can I start with six digit 9s? The sixth 9 would come from 59 in the original number, but then only 83 digits of the original number remain, and I need another 86 digits – no good.

Can I start “999998…”? Well, that 8 would come from 58 in the original number, but then only 85 digits of the original number remain, and I still need another 86 digits – close, but not close enough.

Can I start “999997…”? This time that 7 would come from 57 in the original number. I still need another 86 digits, but now I have 87 digits of the original number to choose from – SUCCESS!

All that remains it to delete one digit from “58596061…9899100”. If I delete the leading digit 5, I will be left with an 8 at the front. Otherwise, I will be left with a 5 at the front. So I should delete the leading 5.

Therefore, the biggest number that can possibly remain after deleting 100 digits is

This is a wonderful problem: the answer is not immediately obvious, but comes out after a little bit of thought, and absolutely no high-powered mathematics is required to solve it.

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