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.
of positive integers for which there exist positive integers 
and 
such that
holds for all integers 
(Note that 
denotes the greatest common divisor of integers 
and 
)
such that 
for all 
?
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.
from the set of positive integers to the set of positive integers such that, for all positive integers 
and 
, there exists a non-degenerate triangle with sides of lengths![\[ a, f(b) \text{ and } f(b + f(a) - 1).\]](http://latex.artofproblemsolving.com/4/a/b/4ab97da27e05d0f9e7762f74d0d0bd2dc8b9842f.png)
be integers such that, for each 
, there exists a subset 
with![\[
\sum_{j \in S} a_j \equiv i \pmod{2048}.
\]](http://latex.artofproblemsolving.com/8/b/e/8be39aaf83795f611b1c3fa0d5971905bd6ebdfe.png)
Show that for each 
, there is exactly one 
such that 
is divisible by 
but not by 
.
is the summation notation, for instance, 
, while for the empty set 
, one defines 
.
, 
. Construct a circle with center 
and radius 
, intersecting the extensions of 
and 
at 
and 
respectively. Construct a circle with center and radius 
, intersecting the extensions of 
and 
at and . Let 
be the intersection of 
and 
, 
be the intersection of 
and 
, 
(
) be the intersection of and the circle with center , and 
(
) be the intersection of 
and the circle with center . Show that , , and are concyclic.
be a natural number. At a mathematical olympiad training camp the same courses are organised every day. Each student takes exactly one of the courses each day. At the end of the camp, every student has takes each course exactly once, and any two students took the same course on at least one day, but took different courses on at least one other day. What is, in terms of , the largest possible number of students at the camp?
be natural numbers such that![\[
\frac{a^n - 1}{(a - 1)^n + 1}
\]](http://latex.artofproblemsolving.com/6/2/7/627f1d5f55e7594f2e6dcf5d541e796b7a5bfae2.png)
is a natural number.
is odd.
be a prime divisor of .![\[
a^{(q - 1)/2} \equiv 1 \pmod{q}.
\]](http://latex.artofproblemsolving.com/b/2/1/b2135d8c5526e577722718e6d02bcc52559b0555.png)
3. Prove that if a is prime and 
, then![\[
2^{(a - 1)/2} \equiv 1 \pmod{a}.
\]](http://latex.artofproblemsolving.com/2/d/c/2dc6cc1fe27cc72e72d63b13cf07939e19051245.png)
be a regular heptagon with its center 
. 
is the orthocenter of triangle 
, is the incenter of triangle 
. Let be the midpoint of 
and 
be the intersection point of 
and 
. Assume is the circumcenter of triangle 
. Prove that 
concur at a single point. be a cyclic quadrilateral such that is a diameter of it's circumcircle. Suppose that and intersect at , and 
at 
, 
and 
at 
, and let be a point on . Show that 
is perpendicular to 
if and only if is the midpoint of .
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.
of positive integers, such that every sum
(with 
) is a composite number, but:
for all 
; for all and 
for all 
?
denotes the greatest common divisor of , 
.
of positive integers, such that every sum(with ) is a composite number, but: for all ; for all and for all ? denotes the greatest common divisor of , .
, there exist sequence 
of positive integers such that 
is composite number for all 
and 
for all 
and 
for all 
, we choose sequence 
, then we choose 
where 
is prime number larger than 
and 
is a positive integer such that 
where 
is a prime divisor of 
for all 
will give us for all , 
and 
for all 
, and another bunch of odd primes (pairwise distinct and also distinct from ), say, 
. Now by virtue of CRT, choose a number 
such that:
Now set 
for 
. We claim that this works. Indeed, since the 
's are in an arithmetic progression, we have 
Now 
is always an integer, because by construction, and hence all 's are odd. For 
, the above sum is obviously composite, and for 
, the above sum reduces to 
, which is composite because 
.
, which can't be 
since we've been careful to set 
. The case 
is similar, so we win. 
and so on
then you have that sum is 
which is obviously incorrect since 
is not a composite number.
, then 
, 
and 
. Then inductively construct the 's for 
as follows.
, 
for 
and 
. This can be done by CRT as we can take sufficiently large 
's and 
's to ensure that they are pairwise coprime and also coprime to 
.
consecutive terms of a arithmetic progression whose each term is odd composite number is a composite number.
where n is odd.Clearly the sum ![$ n[x+\frac{n-1}{2}d]$](http://latex.artofproblemsolving.com/d/b/8/db8452ec6c74f166c70507213019065eb9b081c2.png)
is composite.
there is a arithmetic progression with terms such that the following holds:
.Consider the AP,
.
and common difference 
and first terms. devides GCD of 2 terms,
. devides at least one of 
or but in both cases,as 
,
.The 4th line follows from the fact that 
.
by the above 2 claims both (a),(b) holds .
for all 
, where 
.
and similarly 
.
for all , we get 
and since 
we are done!
. (Goal being )
just take 
or anything similar.
we choose primes 
such that they are all distinct and coprime to any subarray sum we have so far. Now we construct 
using 
such that:![\[ a_{n+1} \equiv -\sum_{i=j}^n a_i \mod p_{n,j}^2\]](http://latex.artofproblemsolving.com/a/1/0/a1084b5f9940fd8157deedb9a72c6b71f70534e2.png)
for all 
and![\[ a_{n+1} \equiv 1 \mod a_na_{n-1} \]](http://latex.artofproblemsolving.com/6/7/5/675b5b20c0d78964e79ebd1bf85416a31b9a40bd.png)
This is valid, since each of the squares are coprime to all elements, including 
. Since each new subarray containing is divisible by some square, they will not be prime. Additionally, is coprime to both 
and 
, so the construction is valid for 
and by induction for all , including 2016
and 
Prove that
