Re: Why is 12345678, 85263147, etc. always divisible by 9?

In article <1140377658.936483.30790@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
bill <b92057@xxxxxxxxx> wrote:

Ignacio Larrosa Ca=F1estro wrote:
En el mensaje:1140357815.354141.312390@xxxxxxxxxxxxxxxxxxxxxxxxxxxx,
TonyVarden <TonyVarden@xxxxxx> escribi=F3:
You can arrange the digits from 1 to 8 in any combination & the
resulting number will always be divisible by 9.

ex:
12345678 / 9 =3D 1371742
85263147 / 9 =3D 9473683
etc.
Why?

Don't know the divisibility rule fby 9?

A number is divisible by 9 iif the sum of its digits (in base 10) is
multiple of 9.

But why? If the sum of the digits is a multiple of 7, the number is
not necessarily divisible by 7.

Short answer: because 10 is congruent to 1 modulo 9, but 10 is NOT
congruent to 1 modulo 7.

Long explanation:

If you write a number as (dn)...(d2)(d1)(d0) where d0 is the units
digit, d1 is the tens digit, d2 is the hundreds digit,..., and dn is
the 10^n-s digit, then what you are ->really<- saying is that you are
talking about the number

d0 + 10*d1 + 10^2*d2 + ... + 10^n*dn.

In general, if you have two numbers x and y, and you want to divide
them by a third number z>0, then: if the remainder when you divide x
by z is r and the remainder when you divide y by z is s, then the
remainder when you divide xy by z is the same as the remainder when
you divide rs by z. To see this, simply note that we are saying that

x = zk + r for some integer k
y = zm + s for some integer s

so

xy = z(zkm + ks + mr) + rs

so it is a multiple of z plus rs.

Likewise, the remainder of dividing x+y by z is the same as the
remainder of dividing r + s by z, because

x+y = z(k+m) + (r+s).


Now, because 10 leaves a remainider of 1 when you divide it by 9, the
number 10*d1 will leave the same remainder as 1*d1 does. Likewise,
since 100 leaves a remainder of 1 when divided by 9, the remainder of
dividing 10^2*d2 by 9 is the same as the remainder of dividing d2 by
9. And the remainder of dividing 10^3*d3 by 9 is the same as the
remainder of dividing d3. Etc.

And so the remainder of dividing

d0 + 10*d1 + 10^2*d2 + ... + 10^n*dn by 9

is the same as the remainder of dividing

d0 + d1 + d2 + ... + dn by 9

namely, the sum of the digits.

It also works with 3, because 10 leaves a remainder of 1 when divided
by 3.

If you wanted to do something similar with 7, you would need to note
that 10 leaves a remainder of 3 when dividided by 7; that 100 leaves a
remainder of 2; that 1000 leaves a remainder of 6, that 10000 leaves a
remainder of 4; that 10^5 leaves a remainder of 5; that 10^6 leaves a
remainder of 1; and then we repeat. So you would need to replace

d0 + 10*d1 + 10^2*d2 + ... + 10^n*dn

with

d0 + 3*d1 + 2*d2 + 6*d3 + 4*d4 + 5*d5 + d6 + ...

and check ->that<- number.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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