Shu wrote: Among all eight-digit multiples of four, are there more numbers with the digit $1$ or without the digit $1$ in their decimal representation? It ends with $00$ to $96$ and only $12,16$ have a one. The $d_3$ to $d_7$ have $10$ possible ways. $d_8$ is between $1$ and $9.$ So of all the numbers $\frac{23}{25}(\frac{9}{10})^5*\frac{8}{9}$ don't have a one in their representation of the $225*10^6$ numbers. That number is $\frac{6561*8*23}{25*10^5}$ and is less than half. Hence there are more numbers with the digit $1$ in their representation.

without 1:$23*8*9^5=10865016$ containing 1:$25*9*10^5-10865016=11634984$

If the number of digits goes to infinity, then the fraction of multiples of four without the digit 1 is bounded by a constant (between 1/2 and 2) times $\sum_{1\notin n} \frac{1}{n}/\sum \frac{1}{n}$ which goes to zero, since $\sum_{k=10^N}{10^{N+1}}\frac{1}{k} = \log 10 + o(1)$ and $\sum_{1\notin n} \frac{1}{n}$ converges. Here is my question: For a sequence of digits $S$, what is the smallest integer $n$ such that there are more $n$-digit multiples of 4 with the sequence $S$ than without the sequence $S$?