Project Euler question 41:

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.

What is the largest n-digit pandigital prime that exists?

I have the functions for checking primality, however, instead of going through all numbers up till 987654321, I decided to write a function which generates all the pandigital numbers through recursion.

def make_pan(curr):
    global pandigital_numbers
    if len(curr) < 9:
        for i in range(1, 10):
            if str(i) in curr:
                make_pan(curr + str(i))

There is a list which holds all of the generated pandigital numbers (pandigital_numbers). For some reason, my program is outputting that there is no prime pandigital number in the list. Is there something that my make_pan function is doing to generate all the numbers?

P.S. The initial function call is make_pan("") to start the recursive process


closed as off-topic by Graipher, Stephen Rauch, yuri, Sᴀᴍ Onᴇᴌᴀ, hjpotter92 Aug 13 '18 at 21:01

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  • \$\begingroup\$ Just to be sure: the current code does not produce the correct output? \$\endgroup\$ – Mast Aug 10 '18 at 6:05
  • 2
    \$\begingroup\$ digits 1 to 9 sum up to 45 thus always divisible by 3 \$\endgroup\$ – stefan Aug 10 '18 at 6:24

Stefan is a genius:

Digits 1 to 9 sum up to 45 thus always being divisible by 3.

This got me thinking that if n is equals to the number of digits in a number, then you can have a pandigital number of n = 4 or something which would consist of numbers like 1234 (although it isn't prime)

  • 2
    \$\begingroup\$ By the same logic, it should be noted that \$\sum_{i=1}^8 i = 36 \equiv_3 0\$, which means that the number will have at most 7 digits. \$\endgroup\$ – maxb Aug 10 '18 at 12:10

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