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.
, 
, 
are pairwise distinct real numbers satisfying 
Prove that
Let , , are pairwise distinct real numbers satisfying 
Prove that
, let 
denote the sum of the geometric series ![\[12 + 12r + 12r^2 + 12r^3 + \ldots.\]](http://latex.artofproblemsolving.com/1/3/d/13d279ac0064f9e9937642f21a5ebc096f08b69a.png)
Let 
between 
and 
satisfy 
. Find 
.
and 
be positive integers. is a divisor of 
and is a divisor of 
.
.
and 
be the two foci of ellipse 
. 
is a point on . Let 
be the center of the excircle of 
. When 
, find the minimum eccentricity of .
of natural numbers with 
elements with odd number. A nonempty subset 
of is called 
if the product of the elements of divisible by the sum of the elements of , but not by its square. It is known that the set itself is good. Determine the maximum number of subsets (including itself).
be positive reals with sum 1. Show that for any -tuple of reals 
, there exists an -tuple of reals 
satisfying 
for all 
, where 
.
chessboard such that no two pawns attack each other?
and 
be two positive integers. Prove that the integer![\[a^2+\left\lceil\frac{4a^2}b\right\rceil\]](http://latex.artofproblemsolving.com/1/4/5/14509b500811314c0c363da994e6de875ba468ba.png)
is not a square. (Here 
denotes the least integer greater than or equal to .)
be a triangle with incenter 
, and let 
be a point on line 
satisfying 
. Let the excircle of triangle opposite the vertex 
be tangent to 
at 
. Define points 
on 
and 
on 
analogously, using the excircles opposite 
and 
, respectively.
is cyclic, then 
is tangent to the circumcircle of 
.
such that when 
is written in base 
, its two right-most digits in base are 
.
. 
. First, we take mod 
. 
is the smallest exponent. Then, we take mod 
. 
is the smallest power. Then, we take mod 
. 
. Therefore, the exponent is a multiple of both 
and 
, or 
and 
Prove that
Let 
Prove that
is 
.
is 
,
is the smallest 
so 
already.
is the smallest 
.
then we must have 
is 
and 
,
,
.
is 
and 
.
and 
for 
,
.
,
.
.
except for 
and 
,
,
.
.
cases.
This means 
.
This means 
.
.
by Euler's Totient, so we test 
for 
to see if 
.
do not work, but 
indeed works. In the second case, we have 
,
.
.
do not work. For 
,
,
cannot work. We also have 
upon multiplying the previous equation by 
,
,
doesn't work. But 
indeed works, so the answer is the least common multiple of 
,
.
and 
.
gives 
to be 
but 
.
.
.
.
.
.
and 
,
.
.
and 
.
.
is the minimum possible value of 
.
![\[3^n=(\text{sum of a lot of }143\text{ powers with at least 2 in the exponent})+1.\]](http://latex.artofproblemsolving.com/f/2/8/f287e3c8ad3220b107566c94121c393eb271b1b5.png)
In other words, the power of 
is at least 
,
We work on the first one.
,
,
.
,
.
.![\[\nu_{13}(3^n-1)=\nu_{13}(27^k-1)=\nu_{13}(27-1)+\nu_{13}(k)=1+\nu_{13}(k)\ge 2,\]](http://latex.artofproblemsolving.com/9/6/9/969be16340a112cd9de9d018d6082b0d6c72be1f.png)
so 
.
,
and 
,
.
.
and 
(and 
)![\[3^n \equiv 1 \pmod{11^2}\]](http://latex.artofproblemsolving.com/c/4/e/c4e2b6991c314962c7608b3e009d1bea54984fb5.png)
![\[3^n \equiv 1 \pmod{13^2}\]](http://latex.artofproblemsolving.com/3/4/b/34beba5ab761d2c9deda72049307076d04bbf4dc.png)
;
.
.![\[3^{156} \equiv 1 \pmod{13^2}.\]](http://latex.artofproblemsolving.com/1/1/0/110f907b246db7c43e97efed07b993c0ff2c975a.png)
.![\[(3^{78} - 1)(3^{78} + 1) \equiv 0 \pmod{13^2}\]](http://latex.artofproblemsolving.com/a/a/8/aa834528392b769576bf1d52c26f0094aab34cac.png)
,
.
.
.
(this could be obtained via a difference of cubes factorization).
.
and thus 
is 
.
From CRT, we can separate these into. 
and 
the given equivalency holds. Hence, 
then we would need an odd prime to divide 
absurd. If we instead use 
where 
then we can apply LTE.
Hence, 
and the minimum such value is 
.
.
,
.
be the order of 
;
by the Binomial Theorem. Clearly 
is minimal.
.
which immediately gives 
.
First off we want 
.
so 
and it is easily checked 
all work. Next we need 
.
so that 
.
so we need 
or 
.