Euler_Problem-040.b93

v
>01:55+*:55+*:55+*:55+*:55+*:55+*130p v
         v                                           <
>110p120p>:10g920g**`!#v_10g920g**-10g1+10p20g55+*20p^>v          <
  >30g.@               >1-:10g/20g+\      10g%10g\-1-:|>\55+/\1-:#^_v
  $                                   $               >           :^
^_^#:                                 <#                p03*g03%+55$<



// getDigit (value, digit_idx -> digit)  // digit of a single number at position x

 >v          <
:|>\55+/\1-:#^_$55+%
 >           :^



// digitAt (pos -> digit)                // digit of fraction at position x

        v                                           <
110p120p>:10g920g**`!#v_10g920g**-10g1+10p20g55+*20p^>v          <
                      >1-:   10g/20g+ \ 10g%10g\-1- :|>\55+/\1-:#^_$55+%
                                                     >           :^



[1,0] digitcount
[2,0] digitvalue
[3,0] result

---------------------------------------

This one is really great - I came up with an O(log n) algorithm (crazy fast) for determining the n-th digit.  
First I tested it in LinqPad, so here the C# code for the algorithm:

```csharp
public int digitAt(int pos) {
    int digitcount = 1;
    int digitvalue = 1;
    
    // Get DigitCount of current number
    while(pos > digitvalue * 9 * digitcount) {
        pos -= digitvalue * 9 * digitcount;
        digitcount++;
        digitvalue *= 10;
    }
    
    // current number and digit-position in number
    int value = digitvalue + (pos - 1)/digitcount;
    int digit = digitcount - (pos - 1)%digitcount - 1;
    
    return getInternalDigit(value, digit);
}

public int getInternalDigit(int value, int digit) {
    return (value / (int)Math.Pow(10, digit)) % 10;
}
```