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 = 0.1Where 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 trick here is to identify a cycle. The easiest way I see to do this is to memorize the remainders. If we have a remainder that we occurred previously there is a cycle.
Let's consider 1/3. The initial remainder is 10. For ten divided by three the new remainder is again ten (or one times ten). So we have a one-cycle of 3.
def get_cycle_count(nominator, denominator):
from itertools import count
assert(nominator == 1)
remainders = {}
remainder = nominator
results = []
for i in count():
result = remainder // denominator
remainder = remainder % denominator
results.append(result)
if remainder in remainders:
return i - remainders[remainder]
else:
remainders[remainder] = i
if remainder == 0:
return 0
remainder *= 10
assert(get_cycle_count(1, 7) == 6)
assert(get_cycle_count(1, 10) == 0)
assert(get_cycle_count(1, 6) == 1)
This is a simple divison algorithm. The only special thing is the remainder and that we remember when it occurs the first time. If a remainder occurrs for the second time we substract the position and thus have the lenght of the cycle. With this solution we should be efficient enough to brute force.
s = max([(get_cycle_count(1, i), i)for i in range(1, 1000)])
s = s[1]
assert(s == 983)
print(s)