Math PROOF for divisibility of 3 rule

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Dec 7th 2009, 11:42 AM
Chelsea

Math PROOF for divisibility of 3 rule

I have to write a math proof for the following rule:
A number is divisible by 3 if and only if the sum of its digits are divisible by 3.

My teacher gave us an example of a mathematical proof of the divisibility rule for 9 which is the same as 3 - A number is divisible by 9 if and only if the sum of its digits are divisible by 9

t+u=9
u=9-t
N=10t+u
N=10t+9-t
N=9t+9
N=9(t+1)
9 divides N

I do not understand this process. Can anyone show me the proof for the divisibility rule for 3?
Dec 7th 2009, 01:28 PM
TheEmptySet

(Lipssealed)

Quote:

Originally Posted by Chelsea

I have to write a math proof for the following rule:
A number is divisible by 3 if and only if the sum of its digits are divisible by 3.

My teacher gave us an example of a mathematical proof of the divisibility rule for 9 which is the same as 3 - A number is divisible by 9 if and only if the sum of its digits are divisible by 9

t+u=9
u=9-t
N=10t+u
N=10t+9-t
N=9t+9
N=9(t+1)
9 divides N

I do not understand this process. Can anyone show me the proof for the divisibility rule for 3?

First note that every number can be written as the sum of power of 10

i.e.

$387=3\cdot 10^2+8\cdot 10^1+7\cdot 10^0$

Now let $a_i$ be the digits of a number with the above example
$a_1=7,a_2=8,a_3=3$

We can write any finite number as

$a_ka_{k-1}...a_2a_1=\sum_{i=1}^{k}a_i\cdot 10^i$

Now if we reduce this equation remember that

$10^i \equiv 1 \pmod{3}$ for every $i \in \mathbb{N}$

$a_ka_{k-1}...a_2a_1 \pmod{3}=\sum_{i=1}^{k}a_i\cdot 10^i\pmod{3}=\sum_{i=1}^{k}a_i \pmod{3}$

From here you can make the if and only if conclusion.

I hope this helps.
Dec 7th 2009, 01:48 PM
deus87

Quote:

Originally Posted by Chelsea

I have to write a math proof for the following rule:
A number is divisible by 3 if and only if the sum of its digits are divisible by 3.

My teacher gave us an example of a mathematical proof of the divisibility rule for 9 which is the same as 3 - A number is divisible by 9 if and only if the sum of its digits are divisible by 9

t+u=9
u=9-t
N=10t+u
N=10t+9-t
N=9t+9
N=9(t+1)
9 divides N

I do not understand this process. Can anyone show me the proof for the divisibility rule for 3?

It's important to note that your teacher did not show that an arbitrary number is divisible by 9 if the sum of its digits is divisible by 9. He only showed that this holds for the case where the number consist of 2 digits. So for an if-and-only-if proof, you need to do some more effort.

Furthermore, try to explain what is unclear in the 'proof' that your teacher provided?
Dec 7th 2009, 03:34 PM
Chelsea

Honestly, I have no idea what I'm doing I am terrible at math. I just need to know a way to prove that rule of divisibility by 9. I guess she did it that way with a two digit number but apparently thats what she wants.
Dec 8th 2009, 01:26 AM
deus87

Well the only way you are going to get help is if you show some effort. Coming here and asking for the answer won't help you at all and certainly won't improve your math skills.

Obviously if you can proof that a 2 digit number is divisible by 9 if the sum of its digits is divisible by 9, you can also proof that it's divisible by 3. After all, 3*3=9...
Besides that, you can also copy the 'proof' of your teacher and replace the 9 by a 3, but than again, you haven't learned anything.