Project Euler Lab - Problem 74

#74 - Digit Factorial Chains

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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:

\[\begin{align} &169 \to 363601 \to 1454 \to 169\\ &871 \to 45361 \to 871\\ &872 \to 45362 \to 872 \end{align}\]

It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

\[\begin{align} &69 \to 363600 \to 1454 \to 169 \to 363601 (\to 1454)\\ &78 \to 45360 \to 871 \to 45361 (\to 871)\\ &540 \to 145 (\to 145) \end{align}\]

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?

This problem is taken from Project Euler, Problem 74.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=74. Published Friday, 16th July 2004, 06:00 pm. Solved by 30,375 members at time of mirroring.

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