Competition problem — Combinatorics

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Combinatorics Difficulty 5.1 AIME, harder Prove it Philippines

Problem:

The irrational number $0.123456789101112 \ldots$ is formed by concatenating, in increasing order, all the positive integers. Find the sum of the first 2016 digits of this number after the decimal point.

Solution

Solution:

From $0$ to $99$ , there are $10$ occurrences of $0$ to $9$ in the ones place, and ten iterations of $0$ to $9$ in the tens place. Thus, the total digit sum from $0$ to $99$ is $20(45)=900$ .

From $100$ to $699$ , there are $6$ occurrences of $0$ to $99$ , plus $100$ iterations each of $0$ to $6$ (in the hundreds digit), which gives us a digit sum of $100(21)+6(900)=7500$ .

This gives the numbers $700,701,702,703,704,705,706,707,708$ . These have a sum of $99$ .

Combining everything we have, we obtain a total digit sum of $900+7500+99=8499$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.