Pandigital Prime

There's only $1! + 2! + 3! + \cdots + 9! = 409113$ $n$-digit pandigital numbers. This is a small enough number to brute force, but we can easily optimize even further by applying a divisibility rule.

If the digits of a number sum to a multiple of 3, that number is divisible by 3. Since:

• the digits of every 5-digit pandigital number will sum to 15
• the digits of every 6-digit pandigital number will sum to 21
• the digits of every 8-digit pandigital number will sum to 36
• the digits of every 9-digit pandigital number will sum to 45

all of these numbers will be divisible by 3, and therefore not be prime. Consequently, to find the largest $n$-digit pandigital prime, we only need to check 4-digit and 7-digit pandigital numbers.

from itertools import permutations

pandigitals = set()
for n in (4, 7):
    for permutation in permutations(range(1, n + 1)):
        k = sum(10^i * d for (i, d) in enumerate(reversed(permutation)))
        pandigitals.add(k)

Now just sort largest-to-smallest and find the first prime.

for p in reversed(sorted(pandigitals)):
    if is_prime(p):
        break

p

7652413

Relevant sequences

• Pandigital numbers: A352991
• Pandigital primes: A216444

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