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.
, 
, 
, 
be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, 
. Prove that 
for all 
.
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
for every pair 
be a trapezoid inscribed in circle 
, 
. Let 
is the symmetric point of 
across 
, 
intersects at 
. Let 
is midpoint of and 
is orthocenter of triangle 
. On 
take a point 
so that 
. Prove that: 
coincide
be integer numbers 
satisfy that:
. Prove that: are even
is a perfect square number
can’t be any power 
of a positive integer number
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
is given where 
are positive reals. Prove that:
with the following property:
number is a divisor of the number if and only if the number is a divisor of the corresponding three-digit number 
.
and 
need not be different.
and 
are numbers that work, but 
and 
don't work.)
satisfying![\[f(x+f(y))=x+f(f(y))\]](http://latex.artofproblemsolving.com/8/d/a/8daa6bcb3155768ee1ba4bd22beb53a85a6e3923.png)
and 
, with the additional constraint 
.
be positive odd integers. Prove that 
doesn't divide 
.
. I couldn't solve this and was wondering if any of you can find the pattern
. I couldn't solve this and was wondering if any of you can find the pattern
.
, and thus the next term is 
and 
intersect at the point 
.
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
families, there are 
children respectively. If a random child from any of the families is chosen, what is the probability that the child has 
siblings? Express your answer as a common fraction.[/quote]
,
people, exactly 
.
lies on diagonal 
such that 
.
.
has 
,
,
.
.
and 
Find the minimum 
,![\[
1, 2, 5, 40, 1280, ?
\]](http://latex.artofproblemsolving.com/a/b/d/abdadcc9d1511e0bc50630d0764932c6723f6123.png)
![\[
a_n = \frac{381}{8}n^{4} - \frac{1885}{4}n^{3} + \frac{13103}{8}n^{2} - \frac{9313}{4}n + 1115
\]](http://latex.artofproblemsolving.com/3/a/5/3a5bf7857ee63dfefc85efdd2f73062889f38dd4.png)
:![\[
a_1 = \frac{381}{8}(1)^4 - \frac{1885}{4}(1)^3 + \frac{13103}{8}(1)^2 - \frac{9313}{4}(1) + 1115 = 1
\]](http://latex.artofproblemsolving.com/a/4/b/a4b8690544780a4953ff30b7e9bfd650d5c05c64.png)
:![\[
a_2 = \frac{381}{8}(2)^4 - \frac{1885}{4}(2)^3 + \frac{13103}{8}(2)^2 - \frac{9313}{4}(2) + 1115 = 2
\]](http://latex.artofproblemsolving.com/4/0/1/401ac0e6b9e04a3bbb18b93adf3afd46aa401c2a.png)
:![\[
a_3 = \frac{381}{8}(3)^4 - \frac{1885}{4}(3)^3 + \frac{13103}{8}(3)^2 - \frac{9313}{4}(3) + 1115 = 5
\]](http://latex.artofproblemsolving.com/7/f/c/7fc785dbf6d9a0992d3bb5e2011bc4db67ef1c02.png)
:![\[
a_4 = \frac{381}{8}(4)^4 - \frac{1885}{4}(4)^3 + \frac{13103}{8}(4)^2 - \frac{9313}{4}(4) + 1115 = 40
\]](http://latex.artofproblemsolving.com/2/1/b/21b4e2e4d19b6d5c983417f70abf485ad3eb155a.png)
:![\[
a_5 = \frac{381}{8}(5)^4 - \frac{1885}{4}(5)^3 + \frac{13103}{8}(5)^2 - \frac{9313}{4}(5) + 1115 = 1280
\]](http://latex.artofproblemsolving.com/b/d/9/bd9f2528ec68ad7182630c86181261f2eeabad17.png)
:![\[
a_6 = \frac{381}{8}(6)^4 - \frac{1885}{4}(6)^3 + \frac{13103}{8}(6)^2 - \frac{9313}{4}(6) + 1115
\]](http://latex.artofproblemsolving.com/4/a/8/4a8c57a4df9d6246743b76544bd06d424c99c27f.png)
![\[
a_6 = \frac{381}{8} \times 1296 - \frac{1885}{4} \times 216 + \frac{13103}{8} \times 36 - \frac{9313}{4} \times 6 + 1115
\]](http://latex.artofproblemsolving.com/b/a/d/bade9652d9835650aa5c7ae840e9564696946715.png)
![\[
a_6 = 69420
\]](http://latex.artofproblemsolving.com/a/9/c/a9cc1cb8407e92daee7dff378e21ddbde47fd9fd.png)
![\[
\boxed{69420}
\]](http://latex.artofproblemsolving.com/2/f/1/2f16609b47f00089d42077ac2f80a849752c8766.png)
.
= (next Catalan number)
= (next Catalan number)
(next Catalan number), 26 + 14 = 40
(next Catalan number), 1238 + 42 = 1280
