AoPS Academy
![$f:[0,1]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/9/d/0/9d00bb0f11e69022dded4a8baba1b496f66d272e.png)
be a continuous function. Prove that 
is a continuous function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
such that 
, then there exists a 
such that:![\[
F(c) = \frac{1}{c} \int_0^c F(x) \,dx
\]](http://latex.artofproblemsolving.com/8/f/e/8fea4d84bc98b6f4953e19976cfa554945a37ea6.png)
, 
is the greatest prime divisor of 
. Find:
![$f(x)\in C[0,3]$](http://latex.artofproblemsolving.com/8/e/2/8e2d8e1cf446975c3e3aa894610ac8b6371737b4.png)
satisfy 
for all 
and
Show that
be a semigroup with the property that for every 
there is a unique 
such that
. Prove that 
is a group.
because you don't know if it exists. This is exactly what the problem asks to prove: that there exists 
and any element is invertible...
???
and its corresponding 
. Then we have 
. Now ![\[((xx'x)(x'))^2=(xx')^4=(xx')^2=(xx'x)(x')\]](http://latex.artofproblemsolving.com/4/d/a/4dadd57901ca2d0e8ccdbdd1569f1a228a699c9e.png)
together with![\[((xx'x)(x'xx'))^2=(xx')^6=(xx')^3=(xx'x)(x'xx')\]](http://latex.artofproblemsolving.com/c/f/7/cf75f505256de47f0f824ecf989d10a18cf8ce0f.png)
we find 
. Similarly 
. We now have 
, which means that the unique 
corresponding to is .
. Take 
such that 
, then 
, which is again equivalent to 
, which means 
. But we also know 
, so 
, hence 
. By uniqueness, we must have 
. Similarly 
.
hence there is an element 
such that 
for all . This satisfy 
and 
, and 
, hence is a group.
is merely a semigroup. There is no addition (and so no subtraction), there is no 
and there is no 
(or