In a message dated 09/10/2004 02:44:15 GMT Daylight Time,

shawns_email_address@... writes:

As it turns out it's a little more general

They don't even need to be consecutive

pick a k and an n, and whala! it's divisible by 3 even if the digits

of n increase mid number or if k>n...although proving this case might

not be so trivial.

(n)--(n+k)--(n+2k)

k=10

n=8

81828

[snip]

Have you not heard of the elementary-school result that if the sum of the

digits of a number is divisible by 3 then the number itself is divisible by 3?

(Ditto 9.)

n + (n+k) + (n+2k) = 3*(n+k).

whala!

-Mike Oakes

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