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www.mikescher.com/www/statics/euler/Euler_Problem-026_explanation.md

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To calculate the repeating-digit-count we use an algorithm based on the idea of a long division. Here on the example of `1/7`
| Position | Value | Remainder | Note |
|----------|-------------- |---------------|----------------------------------------|
| 0 | 0, | 10/7 | `10/7 = 1` **&** `(10%7)*10 = 30` |
| 1 | 0,1 | 30/7 | `30/7 = 1` **&** `(30%7)*10 = 20` |
| 2 | 0,13 | 20/7 | `20/7 = 1` **&** `(20%7)*10 = 60` |
| 3 | 0,132 | 60/7 | `60/7 = 1` **&** `(60%7)*10 = 40` |
| 4 | 0,1328 | 40/7 | `40/7 = 1` **&** `(40%7)*10 = 50` |
| 5 | 0,13285 | 50/7 | `50/7 = 1` **&** `(50%7)*10 = 10` |
| 6 | 0,132857 | 10/7 | **duplicate remainder -> loop closed** |
**=>** RepeatingDigitCount := `6 - 0` = `6`
We use a 1000-field "array" to remember every remainder we already had - as soon as we reach one that is already in use the digits start repeating itself.
For better understanding here the **FindRepeatingDigitCount(int divisor)** algorithm in pseudo-code:
```
int current = 1;
int position = 0;
while(true) {
current = (current*10) % divisor;
position++;
if (grid[current] != 0) return position - grid[current];
else grid[current] = position;
}
```