# Project Euler

### A Taste of Number Theory

#### Problem 26: Reciprocal Cycles

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle. Find the value of d<1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.

#### The Catch

How to find the length of a recurring cycle of 1/d

#### The Light

Use an algorithm which uses modulus operation to find the length of a recurring cycle:

• Take the modulus 1 % d. Call the result x.
• Times x by 10.
• Take the modulus x % d.
• Keep track of all found values for x.
• Repeat step 2 - 3 until x repeats again or x = 0.

For example, to find the recurring cycle length of 1/8:
1 % 8 = 1
10 % 8 = 2
20 % 8 = 4
40 % 8 = 0 (Stop the algorithm here; 1/8 has a recurring cycle length of 3)

#### The Code

```import java.util.*;

public class Problem26
{
public static void main(String[] args)
{
int count = 1;
int remainder = 0;
int tmp = 10;
int max = 0;
int result = 0;

for(int d = 2; d < 1000; d++)
{
ArrayList list = new ArrayList();
while(true)
{
remainder = tmp % d;
if(list.contains(remainder) || remainder == 0)
{
if(count > max)
{
max = count;
result = d;
}
count = 1;
tmp = 10;
break;
}
else
{
tmp = remainder * 10;
count++;
}
}
}
System.out.println(result);
}
}
```