33! will have way more zeros at the end than 23!, surely more than 3 more.

How many more zeros are there at the end of 33! vs 23!?  Factors contributing a "5" would include 25 (2 fives) and 30 (1 five).  So 33! has 3 more zeros than 23!

So we need only find the last 3 nonzero digits of 23! and then subtract that from 1000.

Begin by making a list of prime factors of each number from 2 to 23, 

then delete any factors of 10 (of which there are none since 10 is not prime), 

then delete equal numbers of 2's and 5's

then sequentially multiply the next prime factor by the running product mod 1000.

finally subtract the final product mod 1000 from 1000.

-----------------------------

def prime_factor(n): 

    """  for integer n, return a list of all the prime factors """

    top = n // 2

    factors = []

    

    for i in range(2,top+1):

        while n/i % 1 == 0:

            factors.append(int(i))

            n = n/i

            if n == 1:

                return factors

    if n != 1:

        factors.append(int(n))

    return factors

def countem(aList, value):

    """ return the number of instances of value in aList  """

    ans = 0

    for val in aList:

        if val == value:

            ans += 1

    return ans

def last3digits(n):

    """ output the last 3 nonzero digits of n!  """

    listOfFactors = []

    for i in range(2,23+1):

        for p in prime_factor(i):

            listOfFactors.append(p)

    listOfFactors = sorted(listOfFactors)

    twos = countem(listOfFactors,2)

    fives = countem(listOfFactors,5)

    toBeDeleted = min(twos, fives)

    for k in range(toBeDeleted):

        listOfFactors.remove(2)

        listOfFactors.remove(5)

    truncatedProduct = 1

    for f in listOfFactors:

        truncatedProduct = (truncatedProduct*f)%1000

    return truncatedProduct

print(1000 - last3digits(23))