Math types: When you total a sum of digits in number and sum divides by 3,

[darkstar]

then the orginal number is divisible by 3?

162 (1+6+2=9; 9 divisible by 3, hence 162 is as well.)

Drving me crazy. It seems like it should be straightforward, but maybe its some longer type proof y'all are good at?

Thanks in advance,

darkstar

[EstimatedProphet]

1. Same thing with 9

which makes sense, since 9 is 3x3

[aePrime]

2. Yes, it's true

Is that what you were asking?

[darkstar]

3. No I'm asking why it's true

and your nic seems promising in this regard...

(Sorry. Meant to type "why" in orginal post)

Nice to meet you :hi:

[aePrime]

4. Yes, I saw what you were asking on a second reading.

I should read more closely.

First we need an introduction to modular arithmetic, since it's not seen by most people. Being a computer scientist, I'm going to do it the CS way, and not the mathematician way.

If we take the integers x and y, and perform a modulo on them: x % y (let % stand for modulus (mod)), it means, "What is the remainder of x / y?"

10 % 5 = 0, since 10 is evenly divisible by 5.
10 % 3 = 1, since if we divide 10 by 3, we have a remainder of 1.

For a number to be divisible by 3, that number modulo 3 must be 0.

Mathematicians would write these as 10 === 0 (mod 5) and 10 === 1 (mod 3) respectively.

Something to note here is that any power of 10 mod 3 is 1.
10 % 3 = 1
100 % 3 = 1
1000 % 3 = 1
...

The decimal system that we all know and love is a sum of weighted powers of 10.

162 = 1 * 10^2 + 6 * 10^1 + 2 ^ 10^0
7823 = 7 * 10^3 + 8 * 10^2 + 2 * 10^1 + 3 * 10^0

This is where the modular arithmetic comes in, and the fact that 10^x % 3 = 1 for any integral power of x.

If we take our number in the broken up decimal way, we can mod each term with three (because of the properties of modular arithmetic). Anything multiplied by a power of 10 is 1, as we've seen, so each term is just multiplied by 1. For 162, your example, 1 % 3 * 1 + 6 % 3 * 1 + 2 % 3 * 1 = 1 % 3 + 6 % 3 + 2 % 3 = 1 + 0 + 2 = 3. 3 % 3 = 0, so we can see that the entire term modulo 3 is 0, and therefore it is divisible by 3.

[darkstar]

5. Wow. Thanx for the effort you

put into that. Better yet, I think I've got it. At any rate, I can bookmark it and refer to it tomorrow at work where we've been kicking this around.

Thanks again, AEP. A real pleasure to meet you.

:toast:

[aePrime]

6. Not a problem

I'm not sure if I made it really clear. If you have any questions, feel free to ask.

Pleasure to meet you too :)

:toast:

[Commie Pinko Dirtbag]

9. A simpler way

"Taking a digit away from the tens and summing it to the units" may be understood as:

First we had 10x + y

Then we have x + y

The difference is 9x, which is divisible by 3. Therefore, if the previous number was divisible by 3, the new number is too. Ditto for 9.

The reasoning works with farther-away places too: x0000y = 10000x + y ; (10000x + y) - (x + y) = 9999x + y

[TheBaldyMan]

7. and that proof is almost identical for the one for modulo 9

WhyI get the impression that a lot of homework is getting solved on DU.
Is it ethical to do someone else's homework ?

[Commie Pinko Dirtbag]

10. That's like asking whether Wikipedia is ethical.

All we're doing here is talk about math.

[Peter Frank]

8. That's as clear as mud to the uninitiated...

Got layman's terms?

[Commie Pinko Dirtbag]

11. See if post #9 helps -nt