[/quote]
,
and 
Prove that
Let 
Prove that
of the triangle 
is extended beyond 
to 
so that 
.
is extended beyond 
to 
so that 
.
,
segments 
and 
are parallel. Angle bisectors of 
and 
meet at 
.
and 
meet at 
.
is cyclic
= 2∠AMP.
| 
with 
are positive integers and 
is prime 
)
![\[
1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6}
\]](http://latex.artofproblemsolving.com/b/5/d/b5d49e25bd8d8a211b72372ff967558e1a8365ce.png)
Find the smallest positive integer 
is divisible by 200.
Let ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
denote the set of natural numbers less than or equal to 
![$f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$](http://latex.artofproblemsolving.com/f/d/f/fdf7b2fcfb960e93d59f3f84cef4372262c69499.png)
be the midpoint of side 
at point 
at 
is tangent to 
,
.
Y by jhu08, Sprites, Adventure10, Mango247
Let 
and 
be three positive integers. Prove that there exist a positive integer 
and a set of 
positive integers 
, such that, for every 
, the 
-ary expansion of is a 
-digit palindrome, and the 
-ary expansion is exactly 
.
proposed by Bojan Basic, Serbia
and 
be three positive integers. Prove that there exist a positive integer 
and a set of 
positive integers 
, such that, for every 
, the 
-ary expansion of is a 
-digit palindrome, and the 
-ary expansion is exactly 
.proposed by Bojan Basic, Serbia
Notice that 
since obviously we can select 
such that 
hence 
and at the 
t
given the new 
which satisfies 
and there is at least one 
with 
.
satisfying the following conditions: (i) each of them is coprime with 
;
for all 
;
is coprime with 
.
,
are distinct large primes. Another option is to choose some large odd 
coprime with 
.
,
represent all residue classes modulo 
such that 
.
.![\[ b_0=B, ~ c_i =\frac{B}{d_i + 1}, ~ b_i=d_ic_i=B-c_i, ~ \text{and}~ N = \overline{xyx}_B = x(B^2 + 1) + yB. \]](http://latex.artofproblemsolving.com/4/e/b/4ebecec757b02cc60687accb7c9e6c636787adbb.png)
We claim that these values satisfy all the requirements; moreover, for every 
the 
we have![\[ N = x(d_i + 1)^2 c_i^2+y(d_i + 1)c_i + x = (x \cdot (d_ic_i)^2 + x)+ (2c_ix + y) \cdot (d_ic_i) + (xc_i^2+yc_i). \]](http://latex.artofproblemsolving.com/a/d/7/ad716d5662296e27857890a4e93e7eb99cd24088.png)
Clearly, 
.
and 
.
is divisible by 
,
.![\[ xc_i + y = \frac{xB + y(d_i + 1)}{d_i + 1}=\frac{(xB + y) + d_iy}{d_i + 1}. \]](http://latex.artofproblemsolving.com/9/5/2/9525dbc187f28b5f2a442b204c60b3b5bf0416fc.png)
The numerator of the last fraction is divisible by 
(due to the choice of 
.
;
.
.
,
.
.
