# In Perms Of

13 is a prime number, divisible only by itself and 1. Perm 13 and you get 31, which is also a prime number. The same is true of 17, 37 and 79. There are only two possible permutations – 2 x 1 – of a two-digit number, so base-10 is terminally permal for two-digit primes:

```13, 31
17, 71
37, 73
79, 97
```

What about three-digit primes? Now there are six possible permutations: 3 x 2 x 1. But base-10 is not terminally permal for three-digit primes. This is the best it does:

```149, 419, 491, 941
179, 197, 719, 971
379, 397, 739, 937
```

Fortunately, we aren’t restricted to base-10. Take a step up and you’ll find that base-11 is terminally permal for three-digit primes (139 in base-11 = 1 x 11^2 + 3 x 11 + 9 = 163 in base-10):

```139, 193, 319, 391, 913, 931 (6 primes) (base=11)

163, 223, 383, 463, 1103, 1123 (base=10)
```

Four-digit primes have twenty-four possible permutations – 4 x 3 x 2 x 1 – and base-10 again falls short:

```1237, 1327, 1723, 2137, 2371,
2713, 2731, 3217, 3271, 7213,
7321 (11 primes)

1279, 1297, 2179, 2719, 2791,
2917, 2971, 7129, 7219, 9127,
9721
```

For four-digit primes, the most permal base I’ve discovered so far is base-13 (where B represents [11]):

```134B, 13B4, 14B3, 1B34, 1B43,
314B, 31B4, 34B1, 3B14, 413B,
41B3, 431B, 43B1, 4B13, 4B31,
B134, B143, B314, B413 (19 primes) (base=13)

2767, 2851, 3019, 4099, 4111,
6823, 6907, 7411, 8467, 9007,
9103, 9319, 9439, 10663, 10687,
24379, 24391, 24691, 24859 (base=10)
```

Is there a base in which all permutations of some four-digit number are prime? I think so, but I haven’t found it yet. Is there always some base, b, in which all permutations of some d-digit number are prime? Is there an infinity of bases in which all permutations of some d-digit number are prime? Easy to ask, difficult to answer. For me, anyway.