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- From: pete@bignode.equinox.gen.nz (Pete Moore)
- Subject: Re: Multiples of 9
- Newsgroups: sci.math
- References: <1992Dec18.165248.10914@vax.oxford.ac.uk>
- Organization: Equinox Networks
- Reply-To: pete@bignode.equinox.gen.nz
- X-Newsreader: TIN [version 1.1 PL8]
- Message-ID: <pete.03kt@bignode.equinox.gen.nz>
- Date: 21 Dec 92 16:51:22 +1200
- Organization: Equinox Networks
- Lines: 53
-
- mingos@vax.oxford.ac.uk wrote:
-
- > I hope I'm posting this into the correct maths group, but I have a problem
- >which has started bugging me, it may be an old chestnut, but I can't find a
- >solution that satisfies me.
-
- > The basic thing is that if you multiply any number by 9, you end up with a
- >number whose digits add up to 9, OR whose digits add up to a number whose
- >digits add up to 9, OR...etc, depending on how big the number is.
- > so, for example, 9x12343 = 111087. 1+1+1+0+8+7 = 18. 1+8 = 9.
-
- > I can figure a real hand waving argument for this, but does anyone have a
- >more definitive proof, or can point me towards one? I'm a chemist by
- >background, so I don't know my way around the mathematical lit. I don't even
- >know which branch of mathematics this is...
-
- It's number theory.
-
- First I'll explain congruences, in case you aren't familiar with them.
- I'll use # for the congruence symbol (nomally written like = but with 3
- horizontal lines instead of 2), i.e. I'll write
- a # b (mod c)
- to mean that a - b is an integer multiple of c. In particular, a#0 (mod 9)
- if and only if a is a multiple of 9.
-
- Some relevant properties of congruences:
- (1) a + b # {a (mod c)} + {b (mod c)} (mod c)
- (2) a * b # {a (mod c)} * {b (mod c)} (mod c)
- (3) 10 # 1 (mod 9) and so 10^k # 1 (mod 9) for any positive
- integer k, by applying (2) repeatedly
-
- Suppose n has k base-10 digits. Let these digits (in order) be a_1, a_2,
- ... a_k
-
- Then n = a_1*10^{k-1} + a_2*10^{k-2} + ... + a_{k-1}*10 +a_k
-
- Using (1)-(4) above, n # a_1 + a_2 + ... + a_k (mod 9) and hence n is
- divisible by 9 (i.e. n#0 (mod 9)) if and only if the sum of the digits of n
- is divisible by 9 (and this sum is divisible by 9 iff the sum of _its_
- digits is divisible by 9, etc).
-
- Example:
-
- let n = 5283 then a_1=5, a_2=2, a_3=8, a_4=3,
- and n = 5*10^3 + 2*10^2 + 8*10 + 3
- so n # 5+2+8+3 = 18 (mod 9), since the powers of 10 are # 1 (mod 9).
- But 18#0 (mod 9), and so 5283 _is_ divisible by 9.
- --
- +------------------- pete@bignode.equinox.gen.nz -------------------+
- | The effort to understand the universe is one of the very few things |
- | that lifts human life above the level of farce, and gives it some |
- | of the grace of tragedy - Steven Weinberg |
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