Digit factorial chains : Problem 74 : Project Euler Python Code
The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
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,871 → 45361 → 871
872 → 45362 → 872
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.78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)
How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
Python Code
def digitFactorialChains():
def fact(digit):
l=1
for i in range(digit):
l=l*digit
digit=digit-1
return l
number = 1
digitfactsum = 0
counter = 0
while number < 10**6:
chain = ()
chain = chain + (number,)
strnumber = str(number)
length = 1
while True:
digitfactsum = 0
for i in range(len(strnumber)):
digitfactsum = digitfactsum + fact(int(strnumber[i]))
if digitfactsum in chain:
break
else:
chain = chain + (digitfactsum,)
length = length + 1
strnumber = str(digitfactsum)
"""print chain,length prints the sequnce with length"""
if length == 60:
counter = counter + 1
"""You wont get bored if you print the loop iteration!"""
print number
number = number + 1
print "chains, with a starting number below one million, contain exactly sixty non-repeating terms:",counter
digitFactorialChains()
def fact(digit):
l=1
for i in range(digit):
l=l*digit
digit=digit-1
return l
number = 1
digitfactsum = 0
counter = 0
while number < 10**6:
chain = ()
chain = chain + (number,)
strnumber = str(number)
length = 1
while True:
digitfactsum = 0
for i in range(len(strnumber)):
digitfactsum = digitfactsum + fact(int(strnumber[i]))
if digitfactsum in chain:
break
else:
chain = chain + (digitfactsum,)
length = length + 1
strnumber = str(digitfactsum)
"""print chain,length prints the sequnce with length"""
if length == 60:
counter = counter + 1
"""You wont get bored if you print the loop iteration!"""
print number
number = number + 1
print "chains, with a starting number below one million, contain exactly sixty non-repeating terms:",counter
digitFactorialChains()
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