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

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;
}