# Do it in a minute

Following are the first 10 terms of the sequence of the 2016-digit numbers with all their digits repeating in ascending (or increasing) order.

0123456789012345678901234567890...123456789012345 123456789012345678901234567890...1234567890123456 23456789012345678901234567890...12345678901234567 3456789012345678901234567890...123456789012345678 456789012345678901234567890...1234567890123456789 56789012345678901234567890...12345678901234567890 6789012345678901234567890...123456789012345678901 789012345678901234567890...1234567890123456789012 89012345678901234567890...12345678901234567890123 9012345678901234567890...123456789012345678901234

What are the last 3 digits of the $$2292016^\text{th}$$ term of the sequence?

Clarifications:

• I have considered the first number as a 2016-digit number as well, the number has one leading zero and then the pattern continues

• Each of the numbers has a repeating part $$\dots 0123456789 \dots$$

Try Part 2 of this problem as well.

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