HongKong 2022-23 IMO Selection Tests

For clearer illustration, we add a comma between two integers as we form $n$ by concatenating the integers from $1$ to $100000$, i.e. we write $$n=1,2,3,4,5,6,7,8,9,10,11,12,\dots ,99998,99999,100000.$$

We count the number of occurrences of $2022$ according to the positions of the commas within these digits.

Case 1: 2022 (i.e. no comma within the digits) The first $2$ occurs either as a ten thousands digit or as a thousands digit. Each occurs $10$ times, giving $20$ occurrences in this case: - $2022$, $20221$, $20222$, $20223$, $20224$, $20225$, $20226$, $20227$, $20228$, $20229$; - $2021$, $2022$, $2023$; $12021$, $12022$, $12023$; $\dots$; $92021$, $92022$, $92023$.

Case 2: 202, 2 The first part $202$ may come from the three digit number $202$, the four digit number $2202$, or a five-digit number starting with $2$ (there are $10$ such numbers), giving $12$ occurrences in this case: - $202$, $203$; - $2202$, $2203$; - $20202$, $20203$; $21202$, $21203$; $\dots$; $29202$, $29203$.

* Case 3: 20, 22 The first part $20$ may come from the three-digit number $220$, four-digit number $2220$, or a five-digit number starting with $22$ (there are $10$ such numbers), giving $12$ occurrences in this case: - $220$, $221$; - $2220$, $2221$; - $22020$, $22021$, $22120$, $22121$; $\dots$; $22920$, $22921$.

It follows that the answer is $20+12+12=44$.