[/quote]
,
,
,
very convex if![\[ \frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|
\]](http://latex.artofproblemsolving.com/3/b/a/3ba7733caf51cc70a090920ca21f8b47aad9b1ec.png)
and 
.
such that![\[
2^a \cdot 3^b + 1 = 5^c.
\]](http://latex.artofproblemsolving.com/0/a/c/0ac1f22162eacf698847bceb33f9a337bd53c44b.png)
satisfying
for all integers 
.
,
means 
applied 
,
means 
applied 
satisfying 
.
such that 
(the number of positive divisors of 
be integers and 
be a prime for which![\[ x^2-3xy+p^2y^2=12p \]](http://latex.artofproblemsolving.com/2/4/c/24cb51eb55144a9a1e0e489d115993496a052e79.png)
.
satisfying 
for all positive integers 
be an isosceles trapezoid with 
.
be the midpoint of 
.
and 
the circumcircles of the triangles 
and 
,
be the crossing point of the tangent to 
with the tangent to 
.
is tangent to 
,
denote the feet of the altitudes from 
,
.
and 
be the circles through 
,
other than 
such that![\[ f(x+f(y))=y+f(x+1),\]](http://latex.artofproblemsolving.com/b/a/3/ba35828951778c0d35f50b5b557796bd2cf9e338.png)
for all 
.
Y by David-Vieta
Let 
denotes the set of integers from 
to 
. A magician with an assistant perform a trick. The magician leaves the hall and the spectator writes a sequence of 
elements on the board from the set . The magician’s assistant looks at them and adds 
more elements from to the existing sequence. After that the spectator replaces three of these 
numbers by random elements of (it is permitted to change them by themselves, that is to not change anything at all, for example). The magician enters and looks at the resulting row of numbers and without error names the original numbers written by the spectator. Find the minimal possible for which the trick is possible.
denotes the set of integers from 
to 
. A magician with an assistant perform a trick. The magician leaves the hall and the spectator writes a sequence of 
elements on the board from the set . The magician’s assistant looks at them and adds 
more elements from to the existing sequence. After that the spectator replaces three of these 
numbers by random elements of (it is permitted to change them by themselves, that is to not change anything at all, for example). The magician enters and looks at the resulting row of numbers and without error names the original numbers written by the spectator. Find the minimal possible for which the trick is possible.
and assistant add 
.
are written in the black board. Spectator replaces at most three place of 
and let 
be the resulting sequence. Magician sees 
and he must guess original 
.
(See the popular notation) which means sending message with length 10 adding 
.
is very hard. At least we need some theory like Reed-Solomon error correction. ![$f\in Z_{239}[t]$](http://latex.artofproblemsolving.com/7/4/2/742c21f0eb1a3f16662f2f1c110500396d27d8fc.png)
of degree 
for which
at 
,
to the right-hand side. The spectator will change three numbers, the magician will comes in and tries all the supposed sets of 13 unchanged numbers. Each he will interpolate the corresponding points with a polynomial
.
