OliForum Matematico

We have a total number $ m \geq 2 $ We must find the smallest total number n which is:
$ n \geq m $
and for which:

for all divisions of the set {m, m+1, ..., n} into two subsets at least one of the subsets contains such number a, b, c (not necessary different)
that ab = c

m, n are integers or reals?

reals but total {-1,-2,1,5, etc.}

The result is n=32? If you confirm, i will write the solution.

I have the same but i'm not sure if it's correct

Ok let's write the (hoped) solution.
We have that n=c; in fact, if n>c, we can have a set of {m,m+1,...,c} that can also be divided into two subset as requested. So to find the smallest n we must suppose n=c.
In both subsets we have numbers $ a_1,a_2 $ and $ b_1,b_2 $ such as $ a_1a_2=b_1b_2=c $. Infact if we haven't those numbers in both subsets, we can place c in one of them such as a and b are in one subset and c in the other one.
Now we have that b_1 or b_2 can't be in the first subset because if they can, we can place one of them in the subset and the product will change. Then we have that
$ a_1,a_2,b_1,b_2 $ are all related to a same prime number; in fact, if they aren't, we have that there is a division in two subset such as $ a_1a_2 $ is different from $ b_1b_2 $
So we have that $ a_1,a_2,b_1,b_2 $ are:
$ a_1=a $
$ a_2=a^4 $
$ b_1=a^3 $
$ b_2=a^2 $
and $ c=a^5 $.
So the minimum value are: $ a=2 $ and $ n=c=32 $.

timothy6 ha scritto:reals but total {-1,-2,1,5, etc.}

Just to make easier understand each other, these are called "integer numbers" or "integers" in english, not "total numbers".

Now I know this solution is wrong. Cause n must be related to m (which is given)
So if someone has any solution please post it!