[/quote]
,
such that 
is an integer for every positive integer 
,
be all the positive integers smaller than 
such that![\[gcd(N, c_i + c_{i + 1}) \neq 1\]](http://latex.artofproblemsolving.com/1/0/7/1075290d442e8c1f54237aad8ad07126f455f1f1.png)
for all 
.
of the triangle 
is extended beyond 
to 
so that 
.
is extended beyond 
to 
so that 
.
,
,
,
be positive real numbers such that 
.![$$\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+a^3}{2}}\leqslant a^2+b^2+c^2.$$](http://latex.artofproblemsolving.com/5/a/b/5ab64bd6c09363206f936a4a49a9ac7968218600.png)
be all the positive integers smaller than 
for all 
satisfied 
Find minimum and maximum of 
of positive integers is called central if for every positive integer 
,
terms of the sequence is equal to 
,
,
,
of positive integers such that for every central sequence 
,
,
,
.
and 
.
and 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
and 
.![\[
\frac{1}{x} + \frac{1}{y} + 2 \sqrt{\frac{2}{x^2 + y^2}} + 4 \geq 4 \left( \sqrt{\frac{2}{x^2 + 1}} + \sqrt{\frac{2}{y^2 + 1}} \right).
\]](http://latex.artofproblemsolving.com/1/b/8/1b8ab2e343a3043271f5a392d32b12aca748f1b9.png)
.
,
,
are concurrent and the point of concurrence is 
,
is written in base 
,
.
for Girls, PRISMS Online Math Meet for Girls, is happening this spring! 
Y by Adventure10, Mango247
So I scored way lower than I should have on the AIME I. I probably got a 5~6.
I made some seriously dumb mistakes. I had the ingenious idea of counting all the possible ways in #2, (where the problem simplified to b+c+d = mult. 5) and miscounted because I forgot one case. The worst thing is I checked my counting 2-3 times and came up with the same answer. I also missed #6 because I did 3!*9!*3/12! instead of (3!*9! + 4!*9! + 5!*9!)/12!. I'm getting seriously demoralized, as next year is my last year that I'm eligible for JMO qualification (I also forgot to multiply by 3 on an AMC10 Question). My JMO index should've been a 207.5, but it's going to be like 181.5 now :/
How does one avoid things like this? It seems that I'm messing up most on counting questions.
I made some seriously dumb mistakes. I had the ingenious idea of counting all the possible ways in #2, (where the problem simplified to b+c+d = mult. 5) and miscounted because I forgot one case. The worst thing is I checked my counting 2-3 times and came up with the same answer. I also missed #6 because I did 3!*9!*3/12! instead of (3!*9! + 4!*9! + 5!*9!)/12!. I'm getting seriously demoralized, as next year is my last year that I'm eligible for JMO qualification (I also forgot to multiply by 3 on an AMC10 Question). My JMO index should've been a 207.5, but it's going to be like 181.5 now :/
How does one avoid things like this? It seems that I'm messing up most on counting questions.
![\[ \frac{\frac{32}{25}}{4} = \frac{8}{25} \cdot 5 = \frac{8}{5} \cdot 2 = \frac{16}{5} = 21 \quad \boxed{B} \]](http://latex.artofproblemsolving.com/f/0/2/f027177dc0e9e08b5ca05e127fea1ade65c30164.png)
(cf. AMC 12B 2013, Problem 19)
