[/quote]
,
and 
Prove that
is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to 
is such that 
.
is chosen on the shorter arc 
of triangle 
intersects 
be the common point of 
meets 
.
is tangent to 
satisfying
Prove
be a right angled triangle with 
.
is the midpoint of side 
And 
be an arbitrary point on 
.
over 
intersects lines 
and 
,
and 
.
lies on 
.
be a trapezoid with parallel sides 
.
and 
lie on the line segments 
and 
,
.
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points 
and 
are concyclic.
be a positive integer. A board with a format 
is divided in 
(or 
)
positive integers 
such that 
for any 
,
such that 
?
![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
.
,
is the smallest number different from 
which doesn't divide 
.
.
and 
meet the sides 
and 
respectively. Prove that ![\[ AB+AC\geq4DE. \]](http://latex.artofproblemsolving.com/3/a/4/3a49c5f8bd2930adca7bde2601dd045d4bf05ff6.png)
.
and 
lie on the circles 
and 
,
and 
.
and 
are concyclic.
and the recursive relation
for all natural number 
.
Y by Adventure10, Mango247
Alexander has chosen a natural number 
and has written down in a line,and in increasing order,all his positive divisors 
(where 
and 
).For each pair of neighbouring numbers,he has found their greater common divisor.The sum of all these 
numbers (the greatest common divisors) is equal to 
.Find all possible values of 
.
and has written down in a line,and in increasing order,all his positive divisors 
(where 
and 
).For each pair of neighbouring numbers,he has found their greater common divisor.The sum of all these 
numbers (the greatest common divisors) is equal to 
.Find all possible values of 
.
my beloved
.
for all 
,
Thus for the sum of the written numbers is 
everywhere except for one index, where 
.
,
are odd. Thus there are two "consecutive" factors of 
and 
.
,
and 
are also consecutive and different from our original pair. But we have
which becomes clear if we multiply everything by 
,
so equality does not hold here either.
,
,
and 
are consecutive and different from 
is the least prime divisor of 
,
,
as desired. 
only, which works.
,
except in one instance where 
are both odd and 
.
unless 
must be 
;
and 
.
So 
which is an obvious contradiction as 
is odd. Hence 
.
which is 
is bounded by 
The second 
is bounded by 
Continuing in this manner, the sum is bounded by 
Since 
it suffices to check 
Clearly, only 
is a value 
by euclidean algorithm. Then, the sum is at most 
Thus, there exists exactly one value such that 
Hence, 
Thus, 
must also be odd values. Hence, 
However, observe that 
whereas 
where we also lose values. Thus, 
By earlier, we are done.
