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.
= 2∠AMP.
, and an odd prime 
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of that divides each of them, these powers are not all zero, and are all at most . Prove that there exists some 
(which may depend on 
) such that whenever divides an element of this sequence, the maximum power of that divides that element is exactly 
.
be a positive integer. Consider 
points on the circumference of a circle. pairs, and line segments are drawn connecting the points in each pair.
} such that for every 2 different number 
and 
in this set at last in one of the permutation comes exactly after
: 
such that:
convex polygons, determine the maximum possible number of edges in the obtained figure. polygons with 
vertices and 
edges in the resulting figure, then 
.)
write all its divisors in increasing order: 
. such that 
.
be an inscribed quadrilateral. Circles of diameters 
and 
intersect at points 
and 
, and circles of diameters 
and 
intersect at points 
and 
. The circles of diameters 
and 
intersect in two points 
and 
. Prove that the lines 
and 
are concurrent.
be distinct real numbers such that 
Find the value of 
be distinct real numbers such that 
Find the value of 
be distinct real numbers such that 
Find the value of 
for which there exists a sequence 
of positive integers satisfying the following conditions:[list]
for all integers 
for all integers 
the sequence 
is a permutation of
[/list]
.
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
has only postive integers in forms
where 
![$f:[1,b]\rightarrow \mathbb{R},\ f(x)=x+\dfrac{bc}{x}$](http://latex.artofproblemsolving.com/5/6/d/56d1aa809cfad6667896f3ec4e2a1951b13a394c.png)
. As 
, we infer that ![$f(x)\ge f(b)=b+c,\ \forall x\in [1,b]$](http://latex.artofproblemsolving.com/3/9/6/396ded9b7014efff81a94e281d94b95dc71b62ab.png)
; in particular, 
.
, hence 
. It is easy to see that for 
, if 
, then 
. If 
, 
, so 
, which is in contradiction with the fact that 
. Thereby, 
.
. If 
, contradiction; so 
and 
, or equivalently 
and is odd, we get that 
. Furthermore, 
, hence 
. Now 
which together with the fact that 
, we get that 
, so the family of the solutions 
is described by the set
, then using 
, and 
, we get 
, which is clearly impossible. Thus, 
must hold.
and implies 
. Keeping this in mind, we now observe that, 
, implies 
. Thus, 
. Now, note also that, if 
, then 
, and thus, the largest power of 
dividing either 
or 
is exactly . Clearly, 
, and thus, 
. Moreover, if 
, then we again have a contradiction, as 
. Thus, 
. This yields, 
, which brings us, 
. Now, using 
, we immediately obtain 
. This immediately establishes (since is odd), that .
It follows that 
and 
we get 
Now 
and is an odd multiple of 2, we have that 4 must not divide either one of them. It follows that 
or 
But 
So 
Q.E.D.
with odd positive integers 
such that 
and 
. It is easy to see that 
. Further,
Of course, numbers 
are even. If both 
and 
are divisible by 
, then 
are not divisible by , so 
, that's impossible. Similar things happen when both 
and 
are divisible by , so we have two cases:
. So 
, hence 
. Further, 
, therefore 
, and we finally get 
.
. Here 
, hence . Also we have that 
, therefore and again.
and thus . Let
where 
are odd integers. Then, the condition becomes![\[2^{2k-2} - x^2 = 2^{2m-2} - y^2 \Longrightarrow 2^{2k-2} - 2^{2m-2} = x^2-y^2\]](http://latex.artofproblemsolving.com/4/d/d/4ddafb7c4a2ed27635fd75217fca16fcc439da28.png)
.
Note that![\[k\geq v_2(x^2-y^2) = v_2(2^{2k-2} - 2^{2m-2}) = 2m-2\]](http://latex.artofproblemsolving.com/2/2/a/22aa55b0d3cbecc2c70081b403a9f8361021af84.png)
where the first inequality is true because 
odd implies that
, and 
implies that 
. 
with equality when 
.
Note that![\[2^{2k-2} - 2^{2m-2} = x^2-y^2 \leq (2^{k-1}-1)^2 - 1^2,\]](http://latex.artofproblemsolving.com/f/8/b/f8b20fd1b76a3e6c7b65073e836c05410fbc908d.png)
thus 
, so . , and so the 
equality case must hold and we're done. 
.
:, we have: 
Therefore 
or . Note that: 
Note that are odd integers.
have a difference of 
which is an odd multiple of , we must have either: or 
. Notice that: 
which implies . Notice that: 
which implies . Also, we have: 
. Thus: 
Since is odd integer, and we are done.
. So, 
. Note that 
, and 
.
, 
, a contradiction as 
. So, 
and 
. Since 
, 
and 
. Hence, 
, and 
.
, consider a prime such that 
. 
, a contradiction, so 
.
be odd integers such that 
and 
and 
for some integers 
and 
,
or 
, contradiction )