and 
are chosen uniformly and at random from the disk 
.
and 
,
are all real numbers. The probability that
can be written as 
, where 
are positive integers, gcd 
,
is not divisible by the square of any prime. Compute 
.
of them, there exist four having a common point. Prove that the squares can be divided into at most 
groups, such that
,
. . . , 
in the interval ![$[-1,1]$](http://latex.artofproblemsolving.com/3/9/f/39f506ebeeeb72f62147990614a67f4e30b7c751.png)
where the
be the product of all
to the other 
points. Prove that
ad 
be two circles intersecting at 
and 
The tangent like closer to 
touches 
and 
respectively. If 
and 
find 
Prove that
Prove that
Let 
Prove that
Let 
Prove that
.
Y by
3 Pokemon each have they own number to use on a part of a lock. Star the Pikachu has the number 87, Luna the Eevee has the number 92, and Aero the Mew has the number 79. A part of their code is found in the expression 
.
To double check if it is right then it needs to follow
and 
. If it does follow it then you need to find the sum of the coefficients with the code the code itself, but if not replace the sum of the coefficients with the lowest possible sum of coefficients where 
in 
where X, Y, and Z are positive integers more than 0.
.To double check if it is right then it needs to follow
and 
. If it does follow it then you need to find the sum of the coefficients with the code the code itself, but if not replace the sum of the coefficients with the lowest possible sum of coefficients where 
in 
where X, Y, and Z are positive integers more than 0.![\[
S=87, \quad L=92, \quad A=79
\]](http://latex.artofproblemsolving.com/7/0/6/706ecc898a59947cf1ca523ef741e51b150c7660.png)
and![\[
C \equiv (3S + 5L - 2A) \bmod 101.
\]](http://latex.artofproblemsolving.com/b/4/2/b42238e566af9c4d19f415b39b896da125131c4a.png)
![\[
3S + 5L - 2A = 3 \times 87 + 5 \times 92 - 2 \times 79 = 261 + 460 - 158 = 563.
\]](http://latex.artofproblemsolving.com/8/2/a/82ad403299fa3e70d34fabf01e0674b29b710638.png)
Modulo 101:![\[
C \equiv 563 \bmod 101 \equiv 58.
\]](http://latex.artofproblemsolving.com/f/1/1/f11d2ae75bad57c3ffd00ac90361b1de083978a3.png)
![\[
C \equiv 5 \pmod{8} \quad \Rightarrow \quad 58 \mod 8 = 2 \neq 5,
\]](http://latex.artofproblemsolving.com/4/1/3/413dcbdccfcd9a047b97186b268767a8403c5cca.png)
![\[
C \equiv 3 \pmod{11} \quad \Rightarrow \quad 58 \mod 11 = 3,
\]](http://latex.artofproblemsolving.com/f/e/4/fe44f945fdc28f3cb5f06fb10a1118b4d3ff4046.png)
so 
fails.
:![\[
C \equiv 5 \pmod{8}, \quad C \equiv 3 \pmod{11}.
\]](http://latex.artofproblemsolving.com/e/c/2/ec2f2715f16df5e4a98af5a802711f4c4da9b550.png)
.![\[
8k + 5 \equiv 3 \pmod{11} \Rightarrow 8k \equiv -2 \equiv 9 \pmod{11}.
\]](http://latex.artofproblemsolving.com/8/e/b/8eb0aa838eaf924a04d62c2af0219578bf9775b8.png)
![\[
k \equiv 9 \times 7 = 63 \equiv 8 \pmod{11}.
\]](http://latex.artofproblemsolving.com/2/b/2/2b2ec5684ed030fbb1cd071f1b6e3481f205ef96.png)
![\[
C = 8 \times 8 + 5 = 69.
\]](http://latex.artofproblemsolving.com/4/6/7/467e8efb865092b581f8e35cbce291cd1d8a348e.png)
such that:![\[
(XZ + YL - ZA) \equiv C \pmod{101}.
\]](http://latex.artofproblemsolving.com/0/f/6/0f6fad6b8deb448a3a3c37e43ea6fb575fc687ee.png)
![\[
XZ - ZA + 92Y \equiv 69 \pmod{101} \implies Z(X - 79) + 92Y \equiv 69 \pmod{101}.
\]](http://latex.artofproblemsolving.com/9/b/6/9b6a14813ad212803381e42d934c7a33c462893a.png)
(so 
)![\[
92Y \equiv 69 \pmod{101}.
\]](http://latex.artofproblemsolving.com/6/8/a/68a9a601fd5039205765f9f6f250a4ee14b325cc.png)
![\[
Y \equiv 69 \times 56 = 3864 \equiv 26 \pmod{101}.
\]](http://latex.artofproblemsolving.com/3/f/b/3fb9c0fc1c2f2173001a6436d53bf98d9a2f006b.png)
,
.![\[
X + Y + Z + C = 79 + 26 + 1 + 69 = \boxed{175}.
\]](http://latex.artofproblemsolving.com/6/6/5/6651123b5cbfd6455bd7deae1d1f12c4579bdfff.png)
