Strong number program in Python

A strong number is equal to the sum of the factorials of its digits. 145 is a strong number, because 1! + 4! + 5! = 1 + 24 + 120 = 145.

The program

import math

n = int(input("Enter a number: "))

digits = [int(d) for d in str(n)]
parts = [math.factorial(d) for d in digits]
total = sum(parts)

print(" + ".join(f"{d}!" for d in digits), "=", " + ".join(str(p) for p in parts), "=", total)

if total == n:
    print(f"{n} is a strong number")
else:
    print(f"{n} is not a strong number")

Typed in this example: 145. Press Run to type your own.

Enter a number: 145
1! + 4! + 5! = 1 + 24 + 120 = 145
145 is a strong number

How it works

• str(n) gives the digits as text, and int(d) turns each one back into a number: [1, 4, 5].
• math.factorial(d) calculates d!, so the parts are [1, 24, 120].
• The program prints the whole sum, so you can check it by hand. For 123 it prints 1! + 2! + 3! = 1 + 2 + 6 = 9, and 9 is not 123.
• The program is for positive numbers. A minus sign is not a digit, so int("-") would raise a ValueError.

This is similar to an Armstrong number, which uses powers of the digits instead of factorials.

Finding all of them

A digit can only be 0 to 9, so the program can calculate those ten factorials once, store them in a list, and look them up:

import math

DIGIT_FACTORIALS = [math.factorial(d) for d in range(10)]


def is_strong(n):
    return n == sum(DIGIT_FACTORIALS[int(d)] for d in str(n))


print(*[n for n in range(1, 100_000) if is_strong(n)])

1 2 145 40585

There are only four strong numbers. 1 and 2 count because 1! = 1 and 2! = 2. The search can stop below 10 million: 9! is 362,880, so an 8-digit number gives a digit sum of at most 8 × 362,880 = 2,903,040. That has only 7 digits, so no number with 8 or more digits can be a strong number.