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Problem 34: Digit Factorials

Solution

Let $f(n)$ be the sum of the factorial of the digits of $n$. If ${10}^k\le n<{10}^{k+1}$, then $f(n)\le 9!\cdot (k+1)=362880(k+1)$. If $k\ge 7$, then $362880(k+1)<{10}^k$, so $9999999$ is a (loose) upper bound. From there, brute force.

Python Code

from math import factorial
def p34():
	res = 0
	fact = list(map(factorial, range(10)))
	for n in range(3, 10 ** 7):
		n_cpy, f = n, 0
		while n_cpy > 0:
			f += fact[n_cpy % 10]
			n_cpy //= 10
		if n == f:
			res += n
	return res

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