Question

Positive outlook. Do six different positive numbers exist such that their sum equals their product? (Explain your answer!)

Answer 1

i suppose its factors of a perfect number

Answer 2

what do you mean by a perfect number?

Answer 3

no - dont think so

Answer 4

like 6:- 1*2*3 = 1+2+3

Answer 5

I didn't say they had to be integers

Answer 6

- doesnt work for next one 28

Answer 7

ahh!

Answer 8

or in order

Answer 9

Interesting problem.
Building on the initial idea that @cwrw238 came up with, lets try and find a sequence of 4 positive numbers that satisfy the conditions where the first 3 are the numbers found by @cwrw238. So we get:
1+2+3+x=1*2*3*x => 6+x=6x => x=6/5

we now repeat for a sequence of 5 to get:
1+2+3+(6/5)+x=1*2*3*(6/5)*x => (36/5)+x=36x/5 => 36+5x=36x => x=36/31

finally we repeat for a sequence of 6 to get:
1+2+3+(6/5)+(36/31)+x=1*2*3*(6/5)*(36/31)*x => x=1296/1141

so our final sequence is:
1, 2, 3, 6/5, 36/31, 1296/1141

sum of sequence = product of sequence = \(9\frac{87921}{176855}\)

you could use the same rule to obtain a sequence of n different positive numbers whose sum equals their product.

Answer 10

as an aside, I found this very interesting little proof for a generalisation of a sequence of 3 numbers using trig identities: http://www.johndcook.com/blog/2008/11/30/tangent-identity/

Answer 11

Nice and unique ansaneer, here's a general proof that this is possible (it's also a bit simpler)
\[a+b+c+d+e+x=abcdex\]\[a+b+c+d+e=abcdex-x=(abcde-1)x\]\[x={a+b+c+d+e \over abcde-1}\]
I'll look at you link in a sec, I'm eating...

Answer 12

ah - yes - a lot simpler - thanks - and enjoy your food :-)

Answer 13

BTW @ansaneer, a few others and I just started a group called Meta-math. You should consider joining, it has more substantial questions like this (including IMO problems which are almost impossible). I'm trying to get enough people to keep the group active; you clearly have what it takes.

And that proof by trig properties is very interesting, thanks, but I think I need to follow it more closely.

Answer 14

*sorry @asnaseer

Answer 15

Meta-math sounds daunting :-(
I follow maths as a hobby because I enjoy problem solving. My day job is as a software engineer. Would I really be worthy to join such a group?

np about spelling MISHTAKE - after all we're all human :-)

Answer 16

I made up the name meta-math, don't let it scare you. Frankly you and I seem to have similar abilities. I just wanted a group that isn't clogged with questions on how to factor, or differentiate basic functions. Just give it a look-see.
The IMO stuff is downright crazy, for people like Zarkon and James, but I am posting questions similar to these, so browse it and if you see on you think you can do, give it a shot. No penalties given :)

Answer 17

OK - great - and thanks for inviting me - I'll give it my best shot :-)
...I almost feel like a real mathematician now :-D

Answer 18

Me too :)

Answer 19

Your simplification could be generalised to:
Let \(S_n=a_1+a_2+...+a_n\), and \(P_n=a_1\cdot a_2\cdot...\cdot a_n\), where \(a_1, a_2, ...,a_n\) are all unique positive numbers.
Now lets assume we have found a sequence of \(k\) positive numbers such that \(S_k=P_k\), we already know one such sequence which is \(S_3=1+2+3=P_3=1\cdot2\cdot3=6\).
With this knowledge, we can find the next sequence as follows:\[S_{k+1}=S_k+a_{k+1}=P_k\cdot a_{k+1}\]but \(P_k=S_k\), therefore:\[S_k+a_{k+1}=S_k\cdot a_{k+1}\quad\implies a_{k+1}=\frac{S_k}{S_k-1}\]