Math question dealing with bases.

Sep 2, 2002
67
0
0
How can I show that if the sum of the digits of a number in base 10 is divisible by 9 then the original number is also divisible by 9? Also, how can I show that if the sum of the digits of a number in base 8 is divisible by 7 then the original number is also divisible by 7? I think I am to use mod.

Thanks.
 

Mucman

Diamond Member
Oct 10, 1999
7,246
1
0
sorry, but most of the folks on here don't do other people's homework.

at least I've given you 2 bumps
 

silent tone

Golden Member
Oct 9, 1999
1,571
1
76
An algorithm to test this for a given number is simple, but I don't remember how to do a proof for something like this.
 

kherman

Golden Member
Jul 21, 2002
1,511
0
0
What do you need to do? A proof? How to convert from base 10 to base 8???? I am confused.
 

Chaotic42

Lifer
Jun 15, 2001
33,929
1,098
126
Originally posted by: Mucman
All your base are belong to us

Heh, if I were more of a nef, I would have posted this.

So are you trying to prove that given a number in base x, which is divisible by (x-1), will always have digits which, when summed, will be divisible by (x-1)?
 

JupiterJones

Senior member
Jun 14, 2001
642
0
0
Theorem: Let b be an integer. Then an integer a is divisible by (b-1) if and only if the sum of digits in its expansion (a)b is divisible by (b-1)

The proof is based on Modulo Arithmetic. We write a=b mod N iff (a-b) is divisible by N or, in other words, when a and b have the same remainder when divided by N. Such a and b are said to be congruent modulo N. From the definition one can easily derive several properties. For example

If a=b mod N and c=d mod N then (a+c)=(b+d) mod N
If a=b mod N and c=d mod N then ac=bd mod N

The criterion of divisibility by 9 follows from the fact that for every n 10n=1 mod 9. This, in turn, is a straightforward consequence of 10=1 mod 9.
 

rob3rt

Member
Jun 7, 2001
114
1
0
You could use modular arithmetic to prove it or just generalize it.

Take the number 387 for example,

=3(100) +8(10) + 7(1)

=3(99+1) + 8(9+1) + 7

=3(99) + 3 + 8(9) + 8 + 7

We already know that 9 divides into 3(99) and 8(9) all we need to do is check that it divides 3+8+7, OR the sum of the digits
 
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