Problem 32 – Pandigital Products

We say that is a pandigital product if and the string “abn” obtained by concatenating the base-10 representations of , , and contains all of the digits 1-9 exactly once (with no 0s). In problem 32 we are asked to find the sum of all pandigital products.

I ended up being not particularly fancy about this problem. Loop through enough values for , and, for each , loop through enough values for , and see if you get a pandigital product with , , and . We know that it is enough for to venture up to , but what about bounds on ? If has 5 digits, then to be a pandigital product, with factors and , the factors must be either 1 and 3 digits, or 2 and 2 digits. But 9*999 and 99*99 aren’t 5 digit numbers, so we know we can stop looking for once we get to 10000. Similar reasoning shows that must have more than 3 digits, and so, in fact, to be a pandigital product, must have 4 digits (and since they must not be 0, and must be distinct, 1234 is the smallest possibility).

The following code seems to get the job done:

def ispandigprod(a,b,n):
    """ assumes a*b==n """
    # consider the digits in the string "abn"
    # remove 0s and duplicates (with filter, set)
    # if you've still got 9 elements, you're pandigital
    return len(set(filter(lambda t:t, map(int, "%s%s%s" % (a,b,n))))) == 9

def solve():
    """ return the sum of the pandigital products """
    ret = 0
    for n in xrange(1234,9876+1):
        a = 1
        included = False
        while(a < n**.5 + 1 and not included):
            if(n % a == 0):
                b = n//a
                if(ispandigprod(a,b,n)):
                    ret += n
                    included = True
            a += 1
    return ret

I’d honestly like a nicer solution for this problem. The loop I have is pretty stupid, and loops through far more choices than is worthwhile. In particular, I’d like to not think about any or that have a digit that is 0. To ignore with a 0 digit, one could replace the “included = False” line (apparently line 10) with

  included = str(n).count("0") > 0

This seems to cut the run-time down by about 20% (or more), so is maybe worthwhile. I still expect this code could be improved by a bit, but I think I’ll stop here for now.

Tags: prob32