Puzzle - RPL and others

[Albert Chan]

(04-27-2021 08:16 PM)3298 Wrote:  Interesting to note: odd bases never have solutions...
Albert, your turn, fill in the blanks.

We can let the number be x, with digits 1 to n, all distinct, in base n, integer n > 1:

x = Σ(d_k * n^k, k = 0 to n-1)
x (mod n-1) ≡ Σ(d_k * 1^k, k = 0 to n-1) ≡ Σ(d_k, k = 0 to n-1)

This explained the shortcut for mod 9 by adding digits, in decimal.

With all digits distinct: x (mod n-1) ≡ n*(n-1)/2

q*(n-1) + r = n*(n-1)/2

We restrict q as integer, such that 0 ≤ r < n-1
With this setup, x divisible by (n-1) is same as test for r = 0.

If n is even, q*(n-1) + r = (n/2) * (n-1) + 0           ⇒ r = 0
If n is odd, q*(n-1) + r = (n-1)/2 * (n-1) + (n-1)/2 ⇒ r = (n-1)/2 ≠ 0