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.
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.
for which 
divides 
.
and 
, 
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that 
for all 
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist.
, a line 
to tangent , and another circle 
disjoint from such that and lie on opposite sides of . The tangents to from a variable point 
on meet at 
and 
. Prove that, as varies over , the circumcircle of 
is tangent to two fixed circles.
be a triangle and let 
be its incircle. Denote by 
and 
the points where is tangent to sides 
and 
, respectively. Denote by 
and 
the points on sides and , respectively, such that 
and 
, and denote by 
the point of intersection of segments 
and 
. Circle intersects segment at two points, the closer of which to the vertex 
is denoted by 
. Prove that 
.
![\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}](http://latex.artofproblemsolving.com/0/c/d/0cd22d36acedbcb10d76d1b884aa35418273ed52.png)
be positive integers. Prove that the expression![\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]](http://latex.artofproblemsolving.com/1/b/7/1b78a2027c3b7156fbb566901d2cfd088cc09f00.png)
is always a positive integer, and determine all possible values it can take.
be a convex pentagon such that 
, 
, and 
. Prove that the perpendicular line from 
to and the line segments and 
are concurrent.
that satisfies: for all different 
, we have 
?
is a positive integer. Call all positive divisors of which are different from 
and beautiful divisors.We call a special number when it has at least beautiful divisors and difference of any beautiful divisors divides as well. Find all special numbers.
is a positive integer. Call all positive divisors of which are different from and beautiful divisors.We call a special number when it has at least beautiful divisors and difference of any beautiful divisors divides as well. Find all special numbers.
is composite with at least two prime divisors, we may write 
for coprime 
. Then 
divides 
but also their gcd is (a common divisor dividing and also divides ), so 
and 
. If 
, then 
which works. If 
, then (as is even as a product of two consecutive integers) 
must divide 
, so 
, corresponding to 
. However, 
does not work with and 
, 
does not work with and 
, 
does not work with and 
, while 
works since all diiferences are 
.
for some prime , where 
from the problem statement. Then 
must divide 
, so 
divides 
, which implies 
. Now 
works (as 
), while 
does not work with 
. are 
.
is composite with at least two prime divisors, we may write for coprime . Then divides but also their gcd is (a common divisor dividing and also divides ), so and . If , then which works. If , then (as is even as a product of two consecutive integers) must divide , so , corresponding to . However, does not work with and , does not work with and , does not work with and , while works since all diiferences are . for some prime , where from the problem statement. Then must divide , so divides , which implies . Now works (as ), while does not work with . are .
cannot be odd, since the difference of any two (proper) divisors is even and hence cannot divide . If 
for a positive integer 
(note that and 
do not have two proper divisors), then from the proper divisors and 
we get 
, so 
. Hence 
, corresponding to 
, which indeed work.
satisfying
Prove
