ABC = A! + B! + C! Solution

'''

I have to find the 3 digits A,B and C, so the number:
ABC = 100A + 10B + C = A! + B! + C!
Example: For A = 2, B = 5, C = 6, we want: 256 = 2! + 5! + 6!, which is not true
Note: NONE OF THE 3 DIGITS (A, B and C) IS EQUAL TO 0

'''

# Factorial Function
def factorial(n):
    if n >= 1:
        return n * factorial(n-1)
    return 1

# MAIN FUNCTION
print("I have to find the 3 digits A,B and C, so the number:")
print("ABC = 100A + 10B + C = A! + B! + C!")
print("Example: For A = 2, B = 5, C = 6, we want: 256 = 2! + 5! + 6!, which is not true")
print("Note: NONE OF THE 3 DIGITS (A, B and C) IS EQUAL TO 0")
print()
print()
counterOfArrangements = 0
threeDigitNumberIJK = 0
sum = 0
solutionList = []
counterWhenIJKBiggerThanSumOfFactorials = 0
for i in range(1, 10):
    for j in range(1, 10):
        for k in range(1, 10):
            counterOfArrangements += 1
            threeDigitNumberIJK = 100 * i + 10 * j + k
            sum = factorial(i) + factorial(j) + factorial(k)
            # IF I FIND AN ARRANGEMENT = SOLUTION OF THE PROBLEM
            if threeDigitNumberIJK == sum:
                print(str(counterOfArrangements) + ":     " + str(threeDigitNumberIJK) + " = " + str(sum) + " , SOLUTION")
                solutionList.append(i)
                solutionList.append(j)
                solutionList.append(k)

            # ELIF - CASE (When (ABC = 100A + 10B + C) > A! + B! + C!)
            elif threeDigitNumberIJK > sum:
                print(str(counterOfArrangements) + ":     " + str(threeDigitNumberIJK) + " > " + str(sum) + " , NOT SOLUTION")
                counterWhenIJKBiggerThanSumOfFactorials += 1
            # ELSE - CASE
            else:
                print(str(counterOfArrangements) + ":     " + str(threeDigitNumberIJK) + " < " + str(sum) + " , NOT SOLUTION")



# Printing out the solutions and the statistics of the survey
# 1. Firstly, I will find out how many solutions are there
# I remember that my solutionList is in the following form:
# solutionList = [solution1I, solution1J, solution1K, solution2I, solution2J, solution2K, ......, solutionNI, solutionNJ, solutionNK]
# The length can be divided with 3 - Solutions are sets of 3 consecutive numbers in the list
print()
print()
step = 3
for i in range(0, len(solutionList), step):
    print("Solution " + str(i+1) + ":   Numbers " + str(solutionList[i]) + ", " + str(solutionList[i+1]) + ", " + str(solutionList[i+2]))
    # print("Because:   " + str(100 * solutionList[i] + 10 * solutionList[i+1] + solutionList[i+2]) + " = " + str(solutionList[i]) + "! + " + str(solutionList[i+1]) + "! + " + str(solutionList[i+2]) + " = " + str(factorial(solutionList[i]))) + " + " + str(solutionList[i+2] + " = " + str(factorial(solutionList[i+1])) + " + " + str(solutionList[i+2]) + " = " + str(factorial(solutionList[i+2])))
    first = solutionList[i]
    second = solutionList[i+1]
    third = solutionList[i+2]
    print(" Because:     " + str(100 * first + 10 * second + third) + " = " + str(first) + "! + " + str(second) + "! + " + str(third) + "! = " + str(factorial(first)) + " + " + str(factorial(second)) + " + " + str(factorial(third)))