0, 0 and 0. Assuming that you are using the whole numbers, and not just natural numbers.

1, 2, and 3
1 + 2 + 3 = 6
1 x 2 x 3 = 6

1, 2, and 3

1, 2, 3

0+0+0=0
0x0x0=0

Actually, there are an infinite number of possibilities:

-1, 0, 1
-2, 0, 2
-3, 0, 3
etc

Well, the first thing I tried (1,2,3) worked. I'm not going to bother trying the rest to make sure they're the only ones; I've got better things to do with the rest of my life...

1,2,3.

3+2+1=6
3x2x1=6

1,2,3=6

also

1,0,-1=0

In fact, any number, the negative of that number, and zero will work.

zero, zero, and zero

but then again theres negative 1, 1 and 0, and then if you think about it, negavitive some number, that same number, only positive, and zero.

1,2 and 3

you should have specified if your just looking for positive whole numbers.

what about zero, zero, and zero.

1 2 3

1+2+3=6
1*2*3=6

0,0,0 or 1,2,3

1, 2 and 3.

If a,b,c are positive integers such that
a+b+c = abc, then

1 = (a+b+c)/(abc) = 1/(bc) + 1/(ac) + 1/(ab).

Hence at least one of the denominators ab, ac, bc must be less or equal 3, that is: must be 2 or 3. (If each of the three fractions on the right hand-side were less than one third, their sum would be strictly smaller than 1.) But the numbers 2, 3 may be factored only in the form 2=1x2, 3=1x3. Hence necessarily either 1, 2 or 1, 3 are among the numbers a, b, c.

Case 1: a=1, b=2. Then 1/(2c) + 1/(1c) + 1/2 = 1, and we obtain c=3.
Case 2: a=1, b=3. Then 1/(3c) + 1/(1c) + 1/3 = 1,
and we obtain c=2.

It follows that the three numbers must necessarily be 1, 2, 3.

I'm so smart!

You're smarter!

Aw, shucks! But I know ONE person I am way smarter than! Kate knows who I mean!

1+2+3=6
1x2x3=6

1,2,3
0,x,-x

good reasoning... but you only cover the case for positive integers. what about fractional numbers and negatives.

OK lets consider this a bit further...

abc = a+b+c.
abc - c = a+b
(ab -1)c = a+b
c = a+b/(ab-1)

There are limitless possibilities. Just pick any numbers you like for a and b, and work c out.

eg a=2, b=4 c=6/7
eg a=45, b=23, c=68/1034

magical isn't it

Don't know, don't care.

I think the interesting point is really that the numbers are uniquely determined if you consider just positive integers. It is not surprising that there are inifinitely many possibilities if rational numbers are considered. If all integers are allowed, then any triple of the form -a, 0, a is an example. Are there more examples than this type (and 1,2,3 of course) in the case of integers?

Well, -1, -2, -3 (for completeness, but not to forget)

Um! How about:
1+2+3=6
1x2x3=6

Sorry.

1, 2, 3

Theres no positive integer combinations other than 1,2 and 3. Your initial analysis seemed like a complete proof to me. The question therefore is "Is there a combination of negatives and positive integers" that also works.

OK, lets get some thinking caps on here...

Say we have a combination (-a,-b,-c) that works.
(a,b,c) will also work as the total will be as before, but multiplied by -1, and same for the product. We know that the only positive solution is 1,2,3 so from this we can deduce that the only combination of 3 negatives that works is the inverse of (1,2,3) or (-1,-2-3)

So that leaves us with a combination of 1 negative and 2 positives, or 2 negatives, and 1 positive.

Consider the original equation abc=a+b+c

From this we got c = (a+b)/(ab-1)
Or we could have got a = (c+b)/(cb-1)
or b = (a+c)/(ac-1)

If two of the numbers are positive integers, the unknown third MUST be positive, based on one of the 3 equations
If two of the numbers are negative, the third MUST be negative, based on one of the 3 equations
ie theres NO mixed positive negative cobination that will work.

2x2x1
2+2+1

3x2x1
3+2+1

?
?

*faint*

Except that mixed positive-negative answers work when the third number is 0.

Now go on to analyze the case when complex numbers are allowed rather than just integer, or rational, or real numbers.

thats allright. I cant believe no one noticed in qual.

complex numbers?

are those the ones that have "i" in them, as in the square root of negative one.

Complex numbers... hmmmmmm.

A quick try at 1+i, 2+2i, 3+3i gives product = 12i-12, sum = 6+6i.

Tricky, I probably could work it out, but have too much else to do right now.

Yup, those are complex numbers.

You should be able to use the equation you've already derived: c = a+b/(ab-1).

So if a=1+i and b=1-i then c=2 or if a=1+i and b=1+i then c=0.4-1.2i. It seems like it'd be interesting to graph this somehow but I think you'd end up needing to show a surface in four dimensions.

I'll ahve to believe you on that one. I can't remember how to divide complex numbers. eg what is 2/(2i-1)

I have a calculator that can operate on complex numbers so I just took the easy way out and 2/(2i-1)=-0.4-0.8i.

However, a little playing around and I've figured out that dividing complex numbers isn't all that hard. You just multiply top and bottom by the complex conjugate (that is, negate the imaginary part) of the denominator and you'll end up with a real number in the denominator and can then it's just simple algebra.

0 0 and 0