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(); list.add(new Integer(1)); while(true) { remainder = tmp % d; if(list.contains(remainder) || remainder == 0) { if(count > max) { max = count; result = d; } count = 1; tmp = 10; break; } else { list.add(new Integer(remainder)); tmp = remainder * 10; count++; } } } System.out.println(result); } }