Divide it!

Let $n$ be an integer. If $n = \sum_{i=0}^N a_i10^i$ where $a_i \in \{0,\dots, 9\}$ , then we know that $n \equiv \sum_{i=0}^N a_i10^i \equiv \sum_{i=0}^N a_i1^i \equiv \sum_{i=0}^N a_i \pmod 3.$ This means that the remainder of $n$ when divided by $3$ is the same as the remainder of the sum of the digits of $n$ when divided by $3$ . The numbers $123,456,789,101112,\dots,979899$ are all divisible by $3$ since the remainder of the sum of the digits of each of these numbers when divided by $3$ is the sum of the remainder of the digits of a number which is congruent to $1 \pmod 3$ another which is congruent to $2 \pmod 3$ and a third which is divisible by $3$ : $123 \equiv 1+2+3\equiv 1+2+0\equiv 0 \pmod 3 \\ 456 \equiv 4+5+6\equiv 1+2+0\equiv 0 \pmod 3\\ 789 \equiv 7+8+9\equiv 1+2+0\equiv 0 \pmod 3\\ \dots\\ 979899 \equiv 97+98+99 \equiv 1+2+0\equiv 0 \pmod 3.$ By juxtaposing all these numbers together, the number $123456789\dots 979899$ is also divisible by $3$ since the sum of the digits of this number is the sum of the digits of all the numbers above $123456789\dots 979899 \equiv 123 + 456 + 789 + \dots + 979899 \equiv 0 + 0 + 0 + \dots + 0 \equiv 0 \pmod 3.$ By finally juxtaposing $100$ , the number $123456789\dots 9899100$ is congruent to $1 \pmod 3$ because $123456789\dots 9899100 \equiv 123456789\dots 9899 + 1 + 0 + 0 \equiv 0 + 1 + 0 + 0\equiv 1 \pmod 3.$ Finally, this number is even and of the form $3n+1$ so by the chinese remainder theorem, this number is congruent to $4 \pmod 6$ .

The number 123456789......9899 is divisible by 3 (by summing the digits). Therefore, 123456789........9899099 is also divisible by 3 .This implies that 123456789.......9899100 is a number of type 3n + 1 (euclid's division lemma).

Now, 123456789......9899100 could be of type 6n, 6n + 1, 6n + 2, 6n + 3, 6n + 4 or 6n + 5.

Since the given number is even therefore it would be of type 6n, 6n+2 or 6n+4

The given number on divided by 3 gives remainder as 1 which is only possible for 6n +4.

This gives that on dividing 123456789.....9899100 by 6 gives remainder 4.