AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
he may move to any of the points 
, 
, 
, 
and 
. Show that if he starts at 
he can never get to 
.
be the prime numbers listed in increasing order, and let 
be a real number between 0 and 1. For positive integer 
, define![\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in] {\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}} & \mbox{if} \; x_{k-1} \neq 0, \end{cases} \]](http://latex.artofproblemsolving.com/e/1/7/e17b3fc5962fba938dee877dac266c3cd024a6e3.png)
denotes the fractional part of 
. (The fractional part of is given by 
where 
is the greatest integer less than or equal to .) Find, with proof, all satisfying 
for which the sequence 
eventually becomes 0.
of positive integers, Ali proceed the following algorythm: In the i-th step he markes all rational numbers in the interval ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
which have denominator equal to 
. Then he write down the number 
equal to the length of the smallest interval in which both two ends of that is a marked number. Find all sequences with 
and such that for all 
we have
be an acute triangle (
) with circumcircle 
. Point 
is chosen on arc 
, not containing 
, so that, letting 
intersect at 
, one has 
. Points 
and 
lie on lines 
and 
respectively, such that 
is a parallelogram. Point 
is chosen on arc , not containing 
, such that 
. Line 
intersects 
at 
, and line 
intersects 
at 
. Line 
intersects , and 
at points 
, 
and 
respectively.
.
such that for all 
we have:
+1 w
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
-term sequence 
in which each term is either 0 or 1 is called a binary sequence of length . Let 
be the number of binary sequences of length containing no three consecutive terms equal to 0, 1, 0 in that order. Let 
be the number of binary sequences of length that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that 
for all positive integers .
points are given on a circle. On a player's turn, that player must draw a triangle using three unused points as vertices, without creating any crossing edges. The first player who cannot move loses. If Elmo's clone goes first and players alternate turns, who wins? (Your answer may be in terms of .) greater than 10. Then, she multiplies all the digits of and obtains an odd number. Find all possible values of the units digit of .
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
satisfy 
. Prove that
When does the equality occur?
be the incircle of the triangle and with centre 
. Let 
be the circumcircle of the triangle 
. Circles and 
intersect at the point and . Let 
be the intersection of the common tangents of the circles and 
. Show that the circumcircle of the triangle 
is tangent to the circumcircle of the triangle .
and not 
?
and the subset of multiples of 
or 
it follows that 
. To see this value is always attainable, write 
for 
and let 
be the union of all 
for 
- then for 
by Pigeonhole one of the tuples has at least 
elements in , while for 
it is either again this, or contains 
consecutive integers from 
. In any case, contains a tuple of consecutive integers and hence a triple 
of integers. The latter are pairwise coprime, so we are done.
,
for all 
,
,
,
there are at least 
,
.
,
,
,
are mutually prime.
,
.
,
,
,
,
.
.
.
,
.
are all even.
are not mutually prime, so 
.
-
.
,
are 
odd, then among the subset there are no 3 numbers which are mutually prime.
.
.
implies that 
.
,
.
,
,
.
So 
,
,
,
And 
,
,
numbers 
.
numbers, there are at least 
numbers which are contained in the above 
,