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- Path: sparky!uunet!elroy.jpl.nasa.gov!nntp-server.caltech.edu!mokie
- From: mokie@cco.caltech.edu (Michael L. Brundage)
- Newsgroups: sci.math
- Subject: Re: Multiples of 9
- Date: 21 Dec 1992 11:20:52 GMT
- Organization: California Institute of Technology, Pasadena
- Lines: 21
- Message-ID: <1h49akINNot4@gap.caltech.edu>
- References: <1992Dec18.165248.10914@vax.oxford.ac.uk> <pete.03kt@bignode.equinox.gen.nz>
- NNTP-Posting-Host: sandman.caltech.edu
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-
- Or, in simplest of terms (without congruences), if you take a number written
- in decimal form, i.e.
- n = a0 + a1*10 + a2*10^2 + a3*10^3 + ... + ak*10^k
- you see that this can also be expressed
- n = a0 + a1*(9+1) + a2*(9^2 + 2*9 + 1) + a3*(9^3 + 3*9^2 + 3*9 + 1) + ...
- = a0 + a1 + a2 + ... + ak + 9Q
- where Q is a bunch of garbage that you don't care about.
- But then the left-hand-side (LHS) is divisible by a positive integer whenever
- the right-hand-side (RHS) is also; so n is divisible by 9 whenever the sum of
- its digits is, since 9*Q is always divisible by 9.
- [Or, as the previous reply pointed out, n is congruent to the sum of its
- digits mod 9:
- n # (a0+a1+...+ak) (mod 9) .]
-
-
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