/
prob074-slow.py
49 lines (35 loc) · 1.39 KB
/
prob074-slow.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
"""
The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
1! + 4! + 5! = 1 + 24 + 120 = 145
Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169;
it turns out that there are only three such loops that exist:
169 -> 363601 -> 1454 -> 169
871 -> 45361 -> 871
872 -> 45362 -> 872
It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,
69 -> 363600 -> 1454 -> 169 -> 363601 (-> 1454)
78 -> 45360 -> 871 -> 45361 (-> 871)
540 -> 145 (-> 145)
Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with
a starting number below one million is sixty terms.
How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
"""
from math import factorial as f
answers = {}
def factorial_digits(x):
x = str(x)
total = 0
for c in x:
total += f(int(c))
return total
for i in range(1, 1000000):
if i % 1000 == 0:
print "Done", i
chain = [i]
total = factorial_digits(i)
while total not in chain:
chain.append(total)
total = factorial_digits(total)
#print i, len(chain), chain
answers[i] = len(chain)
print len([k for k in answers.keys() if answers[k] == 60])