Reciprocal cycles
Problem 26
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
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.
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
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
Math Concept:
The fraction of the Prime number P except 2, 5 always produce repeated cycles in fractional part, which are most likely of the factor of P-1 or at most P-1.
According to Fermat's theorem, length of the repeated part is given as follows:
10^n mod P = 1
now, this n can be either P-1 or factor of P-1
All we need to do is start from 1000 downwards looking for prime numbers, which satisfies P-1 length.
NOTE: ELEGANT EXAMPLE OF QUICK SOLUTION IF YOU KNOW MATH :)
def is_prime(n):
i = 2
while i*i <= n:
if n%i == 0:
return False
i = i + 1
return True
def generate_prime_below(n):
l1 = []
for i in range(n,1,-1):
if is_prime(i):
l1.append(i)
return l1
for p in generate_prime_below(1000):
c = 1
while pow(10,c,p) != 1:
c = c + 1
if p-c == 1:
break
print p
ANSWER = 983
Thanks
Would (exchanging c with e for my own purposes)
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e = 1
# Start at e=1 and continue to increment e
while pow(10, e, p) != 1:
e += 1
# Stop when 10^e(mod p) ≡ 1 and e is (p-1) or a factor of (p-1). See Fermat's Little Theorem.
if pow(10, e, p) == 1 and e == p - 1:
break
be more clear? I'm not sure why you wrote p - c == 1.
Actually, it should be a^(p-1)-1=0 (mod p). So, you shouldn't be looking for a factor of p-1.
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