Print all n-digit numbers whose product of digits equals to given product python

Question

enter image description here Display 3-digit prime numbers that have the product of digits equal to a given p-value. Example: For p = 9 the numbers 191, 313, 331, 911 meet the conditions of the problem.

I found a code that does this just to find out their amount but not the product. I need to find out what I change in the program to find out the product, not the amount.

Here is the code to be solved in python:

def findNDigitNumsUtil(n, product, out,index):
    # Base case
    if (index > n or product < 0):
        return
    f = ""
    # If number becomes N-digit
    if (index == n):
        # if sum of its digits is equal
        # to given sum, print it
        if(product == 0):
            out[index] = "\0"
            for i in out:
                f = f + i
            print(f, end = " ")
        return
    # Traverse through every digit. Note
    # that here we're considering leading
    # 0's as digits
    for i in range(10):
        # append current digit to number
        out[index] = chr(i + ord('0'))
        # recurse for next digit with reduced sum
        findNDigitNumsUtil(n, product - i,
                        out, index + 1)

# This is mainly a wrapper over findNDigitNumsUtil.
# It explicitly handles leading digit
def findNDigitNums( n, sum):
    # output array to store N-digit numbers
    out = [False] * (n + 1)
    # fill 1st position by every digit
    # from 1 to 9 and calls findNDigitNumsUtil()
    # for remaining positions
    for i in range(1, 10):
        out[0] = chr(i + ord('0'))
        findNDigitNumsUtil(n, sum - i, out, 1)
# Driver Code
if __name__ == "__main__":
    n = 3
    sum = 9
    findNDigitNums(n, sum)

# This code is contributed
# by ChitraNayal

Answer 1

Disclaimer: this code was written by Github Copilot with minimal supervision.

# Display 3-digit prime numbers that have the product of digits equal to a given p-value.
# Example: For p = 9 the numbers 191, 313, 331, 911 meet the conditions of the problem.

def is_prime(n):
    if n == 1:
        return False
    for i in range(2, int(n**0.5)+1):
        if n % i == 0:
            return False
    return True

def print_3_digit_primes(p):
    for i in range(100, 1000):
        if is_prime(i) and (i % 10) * (i // 100) * (i % 100 // 10) == p:
            print(i)

print_3_digit_primes(9)