for which there are infinitely many positive integers 
is not divisible by 
.
and 
.
Equality holds when 
Equality holds when 
,
.
![$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$](http://latex.artofproblemsolving.com/2/2/e/22e6ec2f98777db80e39333542664e903663bf81.png)
![$$a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$$](http://latex.artofproblemsolving.com/f/c/a/fca343ee701eeff4c309b31beb3426ec96b49b55.png)
Let 
.
![$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$](http://latex.artofproblemsolving.com/6/9/5/695cc35986a391fdc15e4863d361ce373fbbf64c.png)
Y by
Here is a complicated problem of calculation. I'd really like to know how you solve it.
Find the minimum
such that there exists exactly 
different functions 
such that ![$f:[1,5]\rightarrow[1,5]$](//latex.artofproblemsolving.com/6/2/0/620e1621905a4670627d46fcb367e1a8c461bb49.png)
satisfying 
.
Find the minimum
such that there exists exactly 
different functions 
such that ![$f:[1,5]\rightarrow[1,5]$](http://latex.artofproblemsolving.com/6/2/0/620e1621905a4670627d46fcb367e1a8c461bb49.png)
satisfying 
.This post has been edited 1 time. Last edited by XAN4, Yesterday at 12:35 PM
Reason: T-y-P-o
Reason: T-y-P-o
.
in the cycle if and only if 
.
.
.
,
.
full cycles will occur and end at 
,
,
has 2 choices, and 
has 3 choices, 
has 4 choices, and 
has 5 choice, all independent of each other. This leaves 120 possibilities in this case.
.
(1 choice for 
,
(1 choice for 
(If 
.
.
divide 
.
is the minimum, we can ignore the 
case entirely.
,
is equal to 120. Therefore, 
,
or 
,
.
,
works, and is indeed the minimum.
which caused me to miss half of the possibilities. This doesn't bring that case up to 
though so I still think both 201 and 471 are wrong.
though. Maybe for 
are wrong. This only leaves me confused.
does not clearly fall into any of the cases, but works for all 
.
