Since the difference under transposition is divisible by b-1, this follows by the fact that the symmetric group is generated by transpositions. In fact, it’s even generated by adjacent transpositions.
Or, alternatively, by the fact that b[sup]n[/sup]-b[sup]m[/sup] is always divisible by b-1 for any two distinct positive integers n and m, n>m:
b[sup]n[/sup]-b[sup]m[/sup] = (b-1) (b[sup]m[/sup]) (b[sup]n-m-1[/sup] + b[sup]n-m-2[/sup]+…+b+1)
So 10000-10, 1000000-10000, 1000-100, what-have-you are all divisible by 9.
Yes, that’s what I said in my earlier post. To get to what erislover said, you need the fact that these transpositions generate the group of all permutations.
[looks closer]
[smacks self on forehead]
You’re absolutely right. Carry on.
So for any base n other than 10, the same kind of interesting math occurs for multiples of n-1?
Divisors, not multiples.
Yeah, AHunter3, that’s right. For example, 1,000,000 (base 10) in base 8 is 3641100.
Add up the digits: 3+6+4+1+1=17 (base 8).
Add these digits: 1+7=10 (base 8).
And again: 1+0=1.
And so we know the remainder when 1,000,000 (base 10) is divided by 7 (=8-1) is 1.
I just saw ultrafilter’s post on preview, so to be clear: In base n, if you add up the digits of any multiple of (any divisor of) n-1, the result will also be a multiple of (any divisor of) n-1. If the number is not a multiple of (any divisor of) n-1, the sum of its digits will still have the same remainder when divided by a divisor of n-1 as the original number, as my example shows.
John Lennon found the “Number nine, number nine” bit from the EMI tape vaults. It came from an old examination tape. The speaker said the number of each question before reading out the question. Lennon looped the bit where he said “Number nine.” It was said that Lennon had some fascination with the number nine. He went on to sing “#9 Dream.”
In Tantric lore, the number nine is associated with the Goddess, Shakti Devi.