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**rida****Real Member**- Registered: 2006-09-25
- Posts: 839

To find multiples of 3 you need to keep on adding the digits until you have a single digit and if that digit is a 3,6 or 9 the number is a multiple of 3.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Very true. In addition, if the digit is a 9 then the original number is a multiple of 9.

Does 0 count as a multiple of 3?

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

i dont quite get how tihs works...

13 isnt a multiple of 3, and 19 certainly isnt a multiple of 9...

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

luca-deltodesco wrote:

i dont quite get how tihs works...

Lets take an example: 314159265

Add the digits: 3 + 1 + 4 + 1 + 5 + 9 + 2 + 6 + 5 = 36

Add the digits of the resulting sum: 3 + 6 = 9

(You keep adding the digits of the resulting sums until you get a single-digit number.)

9 is a multiple of 3.

∴ 314159265 is a multiple of 3.

Another example: 2718281828

Add the digits: 2 + 7 + 1 + 8 + 2 + 8 + 1 + 8 + 2 + 8 = 47

Add again: 4 + 7 = 11

And again: 1 + 1 = 2

2 is not a multiple of 3.

2718281828 is not a multiple of 3.

And **mathsyperson**, yes: 0 is a multiple of 3. An integer *a* is a multiple of an integer *b* iff *a* = *bc* for some integer *c*; since 0 = 3×0, 0 is thus a multiple of 3.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

13 = 1 + 3 = 4 which is not a multiple of 3, so 13 isn't a multiple of 3.

19 = 1 + 9 = 10 = 1 + 0 = 1. Same.

12 = 1 + 2 = 3

105 = 1 + 0 + 5 = 6

You can do the same thing with any base. 11 is a good one to do as it turn out to be:

But some can be just weird. For example, try doing it for base 7.

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**lightning****Real Member**- Registered: 2007-02-26
- Posts: 2,060

3x0=0

3x1=3

3x2=6

3x3=9

3x4=12

3x5=15

3x6=18

3x7=21

3x8=24

3x9=27

3x10=30

that should be the answer to your question

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

JaneFairfax wrote:

And

mathsyperson, yes: 0 is a multiple of 3. An integerais a multiple of an integerbiffa=bcfor some integerc; since 0 = 3×0, 0 is thus a multiple of 3.

I thought it was, but if that's the case then rida's definition isn't quite right. As well as 3, 6, or 9, it also works when all the digits add to 0. Of course, that only happens when the original number was 0 anyway.

If I've followed Ricky's method correctly, then the formula that you get out of it would be a horrible thing that involved a loop of 6 different coefficients for each digit. It'd probably be easier to just try to divide whatever your number is by 7.

I know an alternate way of finding whether a number is a multiple of 7 that's possibly slightly easier than Ricky's (although just dividing by 7 would probably still be easier).

Take the last digit of your number, double it and take it away from the rest of the number.

For example, for 323274 we would take the 4, double it and take it away from 32327, making our new number 32319.

Reiterating this gives 3213, then 315, then 21.

21 is a multiple of 7, and so 323274 also is.

Why did the vector cross the road?

It wanted to be normal.

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