All 3 digit numbers are written into one long number $N=100101102...999$. Find $N \pmod {210}$. I have no idea how to solve this, but I feel that $210=2\times3\times5\times7$ has something to do with the problem.
 A: Let's solve it explicitly based on JMoravitz's suggestion.
For clarity, let's write out the number with spacing between 3 digits:
$$ N = 100 101 102 103 ... 998 999. $$

We claim the following:
$$ \begin{align} 
N &\equiv 1 \pmod 2, \quad (1) \\
N &\equiv 0 \pmod 3, \quad (2)  \\
N &\equiv 4 \pmod 5, \quad (3)  \\
N &\equiv 2 \pmod 7. \quad (4)  \\
\end{align}
$$
Once $(1)$ through $(4)$ are verified, it is straightforward to see that $N \equiv 9 \pmod 210$ where $210 = 2\cdot3 \cdot5\cdot7.$
Proof:


*

*Note that $(1)$ and $(3)$ are obvious.

*Prove (4) using the divisibility rule for 7 stated in the reference above:

"Form the alternating sum of blocks of three from right to left"

Based on this rule:
$$\begin{align} 
   N &\equiv 999 - 998 + 997 - 996 + ... + 101 - 100, \pmod 7 \\
     &\equiv 1 + 1 + ... + 1, \pmod 7 \\
     &\equiv 450 = 7 \cdot 64 + 2, \pmod 7\\
     &\equiv 2 \pmod 7
\end{align} $$


*Prove (3) using the divisibility rule for 3 stated in the reference above:



"Subtract the quantity of the digits 2, 5, and 8 in the number from the quantity of the digits 1, 4, and 7 in the number. The result must be divisible by 3"

To count the digits correctly, we can rely on the pseudo code below to generate (print) the number N:
for (i = 1; i < 10; i++) {
    for (j = 0; j < 10; j++) {
        for (k = 0; k < 10; k++) { 
           print one three-digit block: i, j, k
        }
    }
}

Thus, 


*

*There are 100 blocks having the same 'i' digit occur in the leftmost digit of the block: 10 'j' values times 10 'k' values. The total counts are 300 blocks with 'i' in the {1,4,7} group and 300 with 'i' in {2, 5, 8}.

*Similarly, there are 100 blocks having the same 'j' digit occur in the middle digit of the block: 10 'i' values times 10 'k' values.  The total counts are 300 in the {1,4,7} group and 300 in {2, 5, 8}.

*There are 100 blocks having the same 'k' digit occur in the rightmost digit of the block: 10 'i' values times 10 'j' values. 
The total counts are 300 in the {1,4,7} group and 300 in {2, 5, 8}.
Thus, the total counts are 900 in the {1,4,7} group and 900 in {2, 5, 8}, which means:
$$ N \equiv 0 \pmod 3$$
