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Topic: Proof that a Number is a Multiple of 3 If the Sum Of Its Digits in
Decimal Notation Is a Multiple Of 3

Replies: 11   Last Post: Dec 31, 2008 6:39 PM

 Messages: [ Previous | Next ]
 Nat Silver Posts: 2,082 Registered: 12/6/04
Re: Proof that a Number is a Multiple of 3 If the Sum Of Its Digits in Decimal Notation Is a Multiple Of 3
Posted: Dec 26, 2008 1:38 PM

Michael Ejercito wrote:

> What is the proof that if a the sum of the digits of a number's
> decimal representation is a multiple of 3, then the number is a
> multiple of 3?

(Note that "is divisible by" and "is a multiple of" mean
the same thing.)

First of all, it's important to understand "linearity":

Suppose integers a, b, c satisfy the relation a + b = c.
If two of the integers are divisible by 3, then the third
integer also is divisible by 3.

(If you have two, then you must have three.)

Proof:
Suppose a and c are divisible by 3.
Then a = 3x and c = 3y.
By substitution, 3x + b = 3y, and
b = 3y - 3x = 3(x - x).
By definition, b is divisible by 3.
A similar argument works for the
other two cases.

Proof of the Divisibility by 3 Test.

We prove the Test for three-digit numbers.

Let x be any three-digit number.
Then x = 10h+10t+u, where
h, t, and u are the hundredth's, tenth's, and unit's
digits respectively (and h is not zero).

So, x = (9h+h)+(9t+t)+u

x = 9(h+t) + (h+t+u)

Since 9(h+t) is divisible by 3, by linearity
x is divisible by 3 if and only if h+t+u is also.

Date Subject Author
12/24/08 Michael Ejercito
12/24/08 Robert Israel
12/25/08 RGVickson@shaw.ca
12/25/08 RGVickson@shaw.ca
12/26/08 Michael Ejercito
12/26/08 Dave Seaman
12/26/08 Denis Feldmann
12/26/08 Nat Silver
12/26/08 RGVickson@shaw.ca
12/31/08 Bill Dubuque
12/26/08 RGVickson@shaw.ca
12/27/08 Richard Tobin