**
True or False?
**

${\color{#3D99F6}{0} \text{ is a multiple of } \color{#D61F06}{3}.}$

True
False
It's undefined

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Basically a is multiple of b if & only if a/b belongs to z-{0} So I think answer is false.

Neel Patel
- 5 years, 7 months ago

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The standard definition of divisibility (in number theory) is

Definition. If a and b are integers and there is some integer c such that a = b · c, then we say that b divides a or is a factor or divisor of a and write b|a.

So 3 trivially divides 0 (as do all integers).

If one were to define "multiple" I can't think of any reason to do so by using divisibility (you can readily modify the above definition to show that multiples can be defined by multiplying), but if you wanted to there is no reason to exclude 0 from the numerator. Remember the integers aren't a field (i.e. 0|0 is totally legit).

Michael Baker
- 5 years, 7 months ago

By using the notation "Z" you have said a/b is an element of all integers. Firstly, a/b isn't an integer it is a quotient, Q. Secondly, Integers all {..., -2, -1, 0, 1, 2,...} so the definition you've used is flawed.

Dominic Parnell
- 5 years, 7 months ago

I strongly, vehemently agree with this, from another angle. For an integer (x) to be a multiple of another integer (y), you must be able to divide the integer (x) by some integer (z) to result in (y). There is nothing you can divide into zero to equal 3. Nothing at all. Therefore 0 is not only not a multiple of three, it's not a multiple of anything. Zero is not a multiple of anything because anything multiplied by zero equals zero. And there is no other possible solution.

Tim Druck
- 5 years, 7 months ago

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That's exactly right - so the answer should be false. In their example of buying trays of avocados they are trying to tell you that nothing would be a multiple - but nothing is nothing so at least to me their example makes no sense.

hollie white
- 5 years, 7 months ago

and I think you meant to say you disagree not that you agree with this.

hollie white
- 5 years, 7 months ago

A múltiple of a number gives as result an int number when you put the "multiple" over that number as you said but look... eg. 3/0 I think it's undefined because that is an infinite :)

Daniel Camacho
- 5 years, 7 months ago

This is why people hate math... This answer is wrong and should be false because you cant define a group of something if it is not there. 0 is a multiple of no interger...

Tony Greene
- 5 years, 7 months ago

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You can't even spell 'integer' mate... Zero is in the timestable of every integer.

Stuart Williams
- 5 years, 7 months ago

in group number theory the number 0 is included in the times tables and in the numbers defined as Z

Dominic Parnell
- 5 years, 7 months ago

Suppose the tray has 2 avocados,then will zero become a multiple of 2?

Sivaraj Rajagopal
- 5 years, 7 months ago

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Everything has a multiple of 0

0x0 0x1 0x2

etc.

0x3 = 0 1x3 = 3 2x3 = 6

Jack Muzza
- 5 years, 7 months ago

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Then how many zeros does it take to get to 3?

Daniel Searle
- 5 years, 7 months ago

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@Daniel Searle – 0 is a multiple of 3 in the sense that all multiples of 3 can be expressed as 3k, k in Z. But, when you ask, how many zeros does it take to get three, you are asking if 3 is a multiple of 0. And we all know that isn't possible because 0 is the only multiple of 0.

Jaybee Penaflor
- 5 years, 7 months ago

@Daniel Searle – I got this wrong at first. The multiple is the product. So zero is a multiple of three only when one of the other factors is zero.

Steven Mellard
- 5 years, 7 months ago

@Daniel Searle – this is talking about multiples of 3...not factors. 0 (to the best of my knowledge) isn't a factor of 3

Kareem Mohamed
- 5 years, 7 months ago

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@Kareem Mohamed – It only happens in one special case. Where 0 = (0)n. The only way 0 is a multiple is when the other factor is 0 also.

Steven Mellard
- 5 years, 7 months ago

In which case the answer should be it is undefined, is not it?

Sivaraj Rajagopal
- 5 years, 7 months ago

It is the other way around. You could say everything is a multiple of zero necause the following is true 0 = n(0). Where in is any number. The problem above is phrased as 3=n(0). There is not a value of n that can make that equation true. To further complicate the problem, 3 is prime.

Steven Mellard
- 5 years, 7 months ago

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@Steven Mellard – I made a mistake here. Zero can be the multiple of three but only in the special condition that zero also be the other factor.

Steven Mellard
- 5 years, 7 months ago

Yes. 0 is a multiple of any number. You can always have zero of somethig..

Lemons come in bags of 5 in Australia... So if I go shopping I can buy 1 bag which = 5 lemons 2 bags which = 10 lemons

Or I can decide I don't want lemons. And buy 0 bags of 5 lemons...which = 0

I seem to be really fixated on fruit shopping :p

Kareem Mohamed
- 5 years, 7 months ago

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Zero of three is not having. So you can't "have zero of anything". You can have "something" but you can not have "nothing".

Eric Belrose
- 5 years, 7 months ago

So if you have nothing of something, and you had nothing of something else and the two nothings are equal then you have any two numbers of nothing equalling each other hence all numbers are equal by law of zero which of course is ridiculous.

Daniel Searle
- 5 years, 7 months ago

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@Daniel Searle – Daniel, I'm not sure what the law of zero is you're referring to, but for your example to hold you would need to divide by zero, which, as you know, is undefined

"if you have nothing of something, and you had nothing of something else" a0, b0

"and the two nothings are equal then you have any two numbers of nothing equaling each other" a0 = b0

"hence all numbers are equal by law of zero" a = b?

you can't get from a0 = b0 to a = b without dividing both sides by zero

Tom O'Keefe
- 5 years, 7 months ago

@Daniel Searle – I can't quite understand your comment Could you perhaps rephrase it algebraically? As a linear equation?

Kareem Mohamed
- 5 years, 7 months ago

@Daniel Searle – Back to my lemon analogy

Let me write it algebraically... L = number of lemons per bag B =number of bags T = total number of lemons

T = L * B

We already wrote that L = 5 So if B = 1

T= 5 * 1 T= 5

if B = 2 T= 5 * 2 T= 10 etc.

if B = 0

T = 5 * 0 T = 0

This is a weird question and I am happy to admit that when I clicked true, I was only 70% certain :P But this was why I chose true.

Perhaps there is a more simple and accurate explanation?

Kareem Mohamed
- 5 years, 7 months ago

This would mean any number is a multiple of three because if you buy 1 bag than thats five lemons if you buy zero bags then thats zerp lemons if you buy 100 bags thats 500 lenons so does that also make 100 a muktiple. This doesn't make sense every number can't be a multiple only numbwr that you can multiply to get to three are multiples and theres nothing you can multiply with zero to get three

Lenny Ortiz
- 5 years, 7 months ago

0 goes into 3 an infinite amount of times which implies that infinity*0=3? That doesn't make sense, this answer is wrong.

Callum Goodall
- 5 years, 7 months ago

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Zero doesn't go into 3 at all because by definition you would have to divide 3 by 0, which is undefined.

Dan Daughtry
- 5 years, 7 months ago

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that is why i believe the answer to not be true, but undefined

Callum Goodall
- 5 years, 7 months ago

You are wrong. You're defining factors, not multiples.

Nathan Fuller
- 5 years, 7 months ago

in math you cannot use infinity in a multiplication like that as if it's a "number"...

Dominic Parnell
- 5 years, 7 months ago

If you multiplied infinity times zero you would still get 0. Zero is zero and there is no number that you could multiply it by to get anything other than zero. Therefore zero is not a multiple of three. Zero is merely a place holder or representation for nothing (or null in my world).

hollie white
- 5 years, 7 months ago

Multiple is by definition a product of any quantity and an integer. But any number multiplied by zero will be zero. And you must then be able to divide the multiple by its factor without a remainder. So the answer to the question goes against two basic principle of Multiplicity and Factorization which makes it nonsense

Jeronnel Barican
- 5 years, 7 months ago

My thought is that it should be undefined, not because of division by 0 which doesn't happen, but by basic logic. If a multiple of three is a number of times we can put things into groups of three.... Then since there are no groups of three possible when you have 0 items, it is not a multiple.

It is not that I have three, 0 times. I have no groups of 3. This 0 is not a defined multiple of three.

Math does have a basis in common sense... But doesn't mean it isn't convention.

Eric Belrose
- 5 years, 7 months ago

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That is just a way of explaining the notion of multiple.If $a=bc$ then $a$ is called a multiple of $b$ and $c$ .So, for $0$ to be a mulitple of $3$ , $0$ must equal to $3b$ ,for some $b$ .Hence $0=3b\rightarrow b=\frac{0}{3}=0$ .Hence $0=3\times 0$ so 0 is a multiple of 3.Real life analogies sometimes fail.Let me give you an example:-3 is a multiple of 3,albeit a negative multiple,but a multiple nonetheless.But how can we put things into groups of $3$ (-1) times in real life?We can't.

Abdur Rehman Zahid
- 5 years, 7 months ago

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Not quite, the example presented here 0 = 3 x 0 does not so much to show that zero is a multiple of three as much as it is an example tha tthree is a multiple of zero.

Steven Mellard
- 5 years, 7 months ago

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@Steven Mellard – I take that back. I was thinking about the factors and not the products. It's true if you include zero as a factor. But that is a special case and may not even apply to multiples.

Steven Mellard
- 5 years, 7 months ago

Actually you can. I can owe three things to five people, thus I have a deficit of 15. So if I've eaten three of them. I give them all back unused but I still lack the last three. I have a shortage (-) multiple of three.

If I get them and give them back my result isn't a multiple of three, it is nothing. The universe is happy. Nothing is no more a multiple of three than five.

Thus other than by convention, 0 is a multiple of nothing.

But I stated that by convention already 0 is a multiple of everything.

I think without convention it is nonsense in the highest order (mathematically) but I already conceded there too in my previous post.

Eric Belrose
- 5 years, 7 months ago

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@Eric Belrose – Of course the shortage is a proxy... But it is something...

Eric Belrose
- 5 years, 7 months ago

3 is a prime number, in which case it only has 2 multiples, itself and one. This means that 0 isn't a multiple of 3.

Linh Vu
- 5 years, 7 months ago

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You've confused factors and multiples. 3 has only two factors, i.e., it itself is a multiple of only two numbers, 1 and 3. But, 3 has an infinite number of multiples, like 0,3,6,9 and so on.

Soumil Aggarwal
- 5 years, 6 months ago

That doesn't make sense at all.

Paul Mears
- 5 years, 7 months ago

Since when are people allowed to use everyday experience to prove math?

Cindy Cin
- 5 years, 7 months ago

That's a bad way of thinking about it. The word multiple also implies a divisor. It would be smarter to say you had a cake and divided it into 3 parts. Then that cake would be a multiple of 3 parts. But if you had 0 cake a divided it into 3 parts you still have no cake. LOL.

Daniel Searle
- 5 years, 7 months ago

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That's a little convoluted way of thinking about it.. but if you take your example and apply it to the definition of a multiple: "b is a multiple of a if b = na for some integer n" Then you have: one "Cake" is a multiple of "Parts" for some integer 3.

Therefore, zero "Cake" is a multiple of "Parts" for some integer 0. ie: no pieces equal no cake.

Or if you had no cake and divided it equally among three people, they would each have no pieces.

Tom O'Keefe
- 5 years, 7 months ago

If you divide 0 cakes into 3 parts then each part has size 0, so 0 is indeed a multiple of 3, and 3 in that sense a divisor of 0. (But in the way mathematicians usually construe the phrase "divisor of zero" it means a nontrivial divisor of zero: you multiply two nonzero things and get zero. This is impossible with numbers but possible with matrices and with congruence classes in modular arithmetic when the modulus is a composite number.)

Michael Hardy
- 5 years, 7 months ago

LOL very nice!

gholamali hosseini
- 5 years, 7 months ago

That's right, but... a múltiple of a number gives as result an int number when you put the "multiple" over that number eg. 3/0 and I think it's undefined because that is an infinite

Daniel Camacho
- 5 years, 7 months ago

Sorry champ but zero is a special number. Find me one other number that you can multiply zero by in order to get a product of three. I've had to explain this to students before. Zero is not just a number it is a concept.

Steven Mellard
- 5 years, 7 months ago

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Whoops I was thinking about factors, not multiples.

Steven Mellard
- 5 years, 7 months ago

Actually this is false, if it where true then 0 would also be a multiple of 1,2,4,5,6,7,8,9,10,11,12,etc.

Dakota Davis
- 5 years, 7 months ago

it makes sense if you say it like that... I was thinking that 0 multiplied by anything is zero....so zero times anything could never equal three. but that explanation helps.

Carter Pugh
- 5 years, 7 months ago

NO. The word "multiple" is not interchangeable with divisibility. A multiple of 3 takes the form 3n, where n is a natural (counting) number. I OBJECT to my answer being marked wrong. I expect this to be corrected.

Pamela Barnes
- 5 years, 7 months ago

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Pamela, can you find a source that says that n has to be a natural number? The common definition states that n has to be an integer.

Tom O'Keefe
- 5 years, 7 months ago

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Thank you for your response.

I taught Algebra 1 and 2 topics for over 30 years. It's what I learned in my teacher education, and EVERY textbook had it. And like I wrote, "multiple" is related to divisibility, but they are not interchangeable. The prime factorization must include a 3 for multiples of 3, and we don't have prime factorizations of negative numbers. (We can write "complete factorizations", by using -1 as the first factor in the factorization.) And we certainly cannot write a prime factorization of zero.

An additional difficulty is the Least Common Multiple, which is necessarily a positive number, because the issue doesn't even come up with respect to negative numbers.

Pamela Barnes
- 5 years, 7 months ago

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@Pamela Barnes – I don't understand what Prime Factorization and Least Common Multiple have to do with the question. The definition of Multiple is:

"A multiple of a number x is any quantity y=nx with n an integer. If x and y are integers, then x is called a factor of y." Weisstein, Eric W. "Multiple." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Multiple.html

So applying this definition to the question, we have: y=0; n=0; x=3 0 = 0*3 n:=0 is an integer Therefore 0 is a multiple of 3

Also, since 0 and 3 are integers, then 3 is a factor of 0. It should be pointed out that Woflram assumes factor to mean integer factor (as the factors include the set of all integers {...-3,-2,-1,0,1,2,3...}. That's probably where the confusion is coming from as in some texts and definitions factors are the set of all positive integers. However, i have yet to come across a definition that contradicts the original question.

Tom O'Keefe
- 5 years, 7 months ago

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@Tom O'Keefe – If I accept that definition of multiple, then zero is a common multiple of 1, 2, 3, 4, 5, ...

Which would make zero the Least Common Multiple of 1, 2, 3, 4, 5, ...

Do you see the problem now?

The definition of multiple must be restricted to natural numbers.

By the way, I checked about 6 online math dictionaries, and of the 3 that had the word multiple, none had a really usable definition. Certainly not a definition that would allow for the difficulty arising from the LCM. ("98 is a multiple of 7 because you can write 7 x 14 to get it." One mentioned a definition almost like mine, not quite as precise, but used "whole numbers", and because of the imprecision, I wonder if the author is using "whole numbers" correctly.)

Pamela Barnes
- 5 years, 7 months ago

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@Pamela Barnes
–
Ahhh, okay, I see your confusion. When you look for the Least Common Multiple, youre evaluating a
*
subset
*
of all multiples. That subset being all positive integers. That doesn't mean that all multiples have to be positive integers.

Tom O'Keefe
- 5 years, 7 months ago

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@Tom O'Keefe – The words Least and Common do not, in themselves, denote positivity, so the word Multiple must denote it.

Pamela Barnes
- 5 years, 7 months ago

@Tom O'Keefe – I don't see how that takes care of the difficulty of zero being the LCM of any group of numbers. The definition of LCM is the multiple of least value that a given group of numbers have in common. There is no additional restriction on the value.

Oh, I did find a definition of multiple that agrees:

http://www.mathgoodies.com/lessons/vol3/lcm.html

"The multiples of a whole number are found by taking the product of any counting number and that whole number."

"Multiple" is just a much more restricted word than many people would like it to be. Which is why my initial answer, False, should be the correct answer. Zero is certainly divisible by 3. But it doesn't follow that zero is then a multiple of 3.

Pamela Barnes
- 5 years, 7 months ago

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@Pamela Barnes – There's really not much more to say. The link you provided shows a list of problem with no definitions. A simple google search will provide a wealth of referencable results that 0 is a multiple of any number. As well as definition of Least Common Multiple. I would suggest you spend some time sifting through the results and re-familiarize yourself with the concepts.

Tom O'Keefe
- 5 years, 7 months ago

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@Tom O'Keefe – Nice. And I was just thinking that I was enjoying our discussion, but you had to blow it. (I pasted the definition from my link. Didn't you see it? And I told you I looked at 6 other online math dictionaries, and only 2 or 3 of those had any definition to look at, and they were inadequate. Not "more like yours". Inadequate.)

The original poster used trays of avocados to argue that zero is a multiple of 3. That clearly indicated two things to me: 1) that the author had confused multiple and divisibility, and 2) that the topic under discussion was NOT at a postgraduate level of mathematics.

The common definition of multiple is as I gave it in a previous response, and I found an online math dictionary which agreed (which I hadn't checked before), in addition to my own 30+ years of teaching math at the level of the original question posed (using many textbooks at that level).

In all common usage of multiple, at the level indicated by the original question and its justification, negatives, fractions, decimals, and zero are not considered multiples of any counting numbers.

For other purposes, the definition can likely be tortured to barely resemble the original meaning, as some of the other posters and you have shown.

But zero is NOT a multiple of 3, at any level of grade school or high school math - the level at which a tray of avocados would be used to illustrate.

Thank you for your condescension, but I assure you it is unneeded. If that is representative of the commentary here, perhaps I'll satisfy my math jones elsewhere. Peace.

(Incidentally, you can't use a tray of avocados to illustrate -3 as a multiple of 3.)

Pamela Barnes
- 5 years, 7 months ago

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@Pamela Barnes – I used to like the "challenges" posted on Brilliant, up until the "official" answer given to this idiotic one. This is NOT a trick question, this is NOT PhD level Math, this is just plain wrong. My way of explaining its wrongness in few words? Common definition of prime number: a number whose only factors are 1 and itself; a number, larger than 1, that can only be divided evenly by itself and 1; a positive integer greater than 1 that can only be divided by itself and 1 without leaving a remainder, with 1 often excluded.

That being said, how is it possible to have prime numbers if 0 is the "third multiple" of every one of them?

Victor Soares
- 5 years, 7 months ago

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@Victor Soares
–
You have that backwards Victor. It should be 0 is the "zeroth multiple" of every number or 0 is a multiple of every number. This would mean that 0 = 0n. In the cases where n is an integer this means that every number (n) is a factor of 0. That's why most (if not all) common definitions of prime number state that it is a number who's only
**
positive
**
factors are 1 and itself. The
**
"positive"
**
condition is missing from your definition. That's probably what is causing you confusion.

Weisstein, Eric W. "Prime Number." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/PrimeNumber.html

To Pamela's point, maybe there are some elementary text books that alter the definition to avoid confusion for grade school children. Or maybe there is some change due to the "common core" teaching that is in the news a lot (you know like that example where 5x3 = 3+3+3+3+3 is true but 5x3 = 5+5+5 is false.). But once you get to HS level math and beyond, there is very few people who wouldn't include 0 as a multiple of every number.

nd

Tom O'Keefe
- 5 years, 7 months ago

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The definition of "multiple" implies an Integer. It's really just semantics. The definition (pulled from wiki) is:

"In mathematics, a multiple is the product of any quantity and an integer. In other words, for the quantities a and b, we say that b is a multiple of a if b = na for some integer n"

I have yet to see a definition that states otherwise.

Tom O'Keefe
- 5 years, 7 months ago

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But it makes solving puzzles a very different thing if a big part of it is knowing what the most common assumptions will be... Why test people's ability to infer assumptions? Just make it explicit.

c stroud
- 5 years, 7 months ago

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Well, it's more like it's just testing you to see if you know the definition of multiple but I get your point. I basically just guessed on my side too. Although if I thought of it as lemons and avocados i could have at least made an inferred guess. Also, i just realized that the title for this question is "0 is always a special case" which is misleading since the reason why this is true is because zero isn't a special case. It fits the definition like any other integer.

Tom O'Keefe
- 5 years, 7 months ago

Finally someone nails it on the head. the question is phrased wrong to obtain a correct answer.

Daniel Searle
- 5 years, 7 months ago

I think they meant is 3 a factor of 0

Tony Tu
- 6 months ago

It is so obviously true.

*
a
*
is multiple of
*
b
*
.

means that,

There exists an integer
*
c
*
, which makes
$b \times c$
become
*
a
*
.

For example, 6 is multiple of 3. In this case
*
c
*
would be 2.
But 1 is not multiple of 3, because there is no integer
*
c
*
that could make 3 to 1 by multiplying it.
In the same sense, 0 is multiple of 3 because there exist a
*
c
*
that makes 3 to 0 by multiplying it, and that
*
c
*
is zero.

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0 is a multiple of 3 because 3 multiplied by 0 is 0. So 0 is a multiple of 3.

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This is quite simple if you don't think about it to hard.

A multiple of a number is any number than can be divided by another number without a remainder.

0÷3=0

Therefore, 0 must be a multiple of 3! (And technically any integer)

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0 is a multiple of 3 as 3*0=0

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0 is a multiple of 3 iff $3 | 0$ .

$3 | 0$ iff there exists an integer n for which 3n = 0 is true.

Since 0 satisfies this equation, and because 0 is an integer, 0 is a multiple of 3.

Thus, the answer is $\boxed{\text{True}}$ .

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All Sets containing avocados or trays consist of subsets that are equal to or greater than 3

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Any whole number that can be multiplied by 3 is a multiple. 3 * 0 = 0

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0 is a multiple of ALL integers.

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no. If you looked at it algebraically x*0= 3 - solve for x. 0=3/x - x is undefined because there is nothing you could divide into 3 to get 0.

hollie white
- 5 years, 7 months ago

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0*3=0. Thus there exists an integer 0 such that 3 divides 0. Hence, 0 is a multiple of 3.

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0 is a multiple of 3. A multiple is result, did you hear me? A multiple is a result of multiplying a number by an integer. 3 multiply by the integer 0 equals zero. Therefore 0 is a multiple of 3.

Robert Morgan
- 5 years, 7 months ago

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Problem Loading...

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You can buy a tray of avocados at my local fruit shop...3 avocados per tray.

1 tray = 3 avocados 2 trays = 6 avocados Etc.

So if I bought 0 trays... That's 0 groups of 3... Which = 0.

Hence 0 is technically a multiple of 3!