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?
Well, it obviously can't have eight or nine digits, because the sum of the numbers 1 to 8 are 36, and 1 to 9 are 45. Therefore, any 1 to 8 pandigital number or any 1 to 9 pandigital number will be divisible by three. (and therefore not prime) Similarly for any 5- or 6-digit pandigital. Since we know that there are four-digit pandigital primes, it remains only to check for a 7-digit pandigital prime, and then a four-digit one.
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Okay, I took two submissions on this because the first time I was iterating the permutations in an incorrect order so I found a 7-digit pandigital prime, but the first one I found wasn't the largest. Once I fixed my iteration, it was good.
This iteration technique might be worth paying attention to. It basically runs a point up and down the number trying to change the value at that index. Whenever it gets a value in any spot not the last, it moves the pointer further down the list of digits - setting a value in the last spot means we have a complete number and should evaluate. Whenever it discovers that it's used all the possible values, it sets the current spot to an out-of-range value and moves the pointer back toward the beginning of the list of digits. The out-of-range value is chosen so that when the pointer moves back down it'll start back at the top of the range of valid values.
Anyway, this is a basic technique for many counting-esque tasks:
change the least significant position that you can, and every time you
change position <n> make sure that positions <n+1>...<end> are
set to the limit you want (max val or min val), depending on the order
you're iterating in.