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and primes 
such that:![\[
x^2 = p - 2 \qquad \text{and} \qquad y^2 = 2p^2 - 2
\]](http://latex.artofproblemsolving.com/c/d/8/cd8425e2bd91139758151b57c7a92f8e062ed00f.png)
has the property for every positive integer 
, its 
term is greater than the mean of the first 
terms, and the sum of its first 
terms is a multiple of . Let 
be the number of such sequences satisfying 
. Compute the remainder when is divided by 
.
has side length 
. Point 
lies on segment 
such that line 
is perpendicular to line 
. Given that 
, the area of can be expressed as 
, where 
and are relatively prime positive integers. Compute 
.
and 
satisfy 
, 
, and 
. If the number of possible ordered pairs 
is equal to 
, compute the remainder when is divided by .
be a triangle. Point 
lies on side 
, point 
lies on side 
, and point 
lies on side 
such that 
, 
, and 
. Let 
be the foot of the altitude from 
to . Given that 
, 
, and ![$[AQPR] = \tfrac{3}{7} \cdot [ABC]$](http://latex.artofproblemsolving.com/e/f/1/ef1f89d7ee2b45faac7265d8e2777ba3918ae16e.png)
, the value of 
can be expressed as , where and are relatively prime positive integers. Compute .
and tells Bob that the three resulting remainders are 
, 
, and , in some order. For how many values of is it possible for Bob to uniquely determine ?
satisfying the system of equations 
The value of 
can be expressed as 
, where 
and 
are positive integers such that is square-free. Compute 
. be the set of all monotonically increasing six-term sequences whose terms are all integers between 
and 
inclusive. We say a sequence 
in is symmetric if for every integer 
, the number of terms of 
that are at least 
is 
. The probability that a randomly chosen element of is symmetric is 
, where and are relatively prime positive integers. Compute ., let 
denote the value of the binary number obtained by reading the binary representation of from right to left. Find the smallest positive integer 
such that the equation 
has at least ten positive integer solutions . be a quadratic polynomial with a positive leading coefficient. There exists a positive real number 
such that 
and 
for 
. Compute 
.
such that 
and 
is a multiple of 
.
be a triangle. Consider the points 
as the feet of the altitudes from 
respectively, and its orthocenter which we suppose is the midpoint of 
. Let 
be the midpoint of , be the midpoint of 
, and 
.
be positive divisors of the number 
such that:
.
squares (Note: 
). The middle column has 
squares. Therefore, we add all the squares and get 
. It could also be done using area in terms of the th figure.
are the first differences, then 
are the second differences, since these are the same it is a quadratic). Quadratics are of the form 
. So, plug in 
in for x, and solve for 
. This gives you 
, 
, 
, and 
consist of , 
, 
, and 
nonoverlapping squares, respectively. If the pattern were continued, how many nonoverlapping squares would there be in figure 
?![[asy]
unitsize(8);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);
draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);
draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);
draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);
draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);
draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);
draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);
draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);
draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);
label("Figure",(0.5,-1),S);
label("$0$",(0.5,-2.5),S);
label("Figure",(9.5,-1),S);
label("$1$",(9.5,-2.5),S);
label("Figure",(19.5,-1),S);
label("$2$",(19.5,-2.5),S);
label("Figure",(32.5,-1),S);
label("$3$",(32.5,-2.5),S);[/asy]](http://latex.artofproblemsolving.com/a/4/7/a47c014550eb652aedf4605a7c120e471994b2fa.png)
, we count the squares with rows of length and 
.
squares. Plug in 
to obtain 
squares in figure 
.
in a continuous form and integrating would yield 
.
to let 
.
is closest to 
and 
.
squares, the second has 
squares, the third has 
squares, and etc.
squares.
squares.
,
squares of one colour and 
squares of the other colour. So, the total is 
.
small squares in this figure.
squares in them. Thus, in total, you're removing 
squares when you remove all four corners.
squares left in the figure.
.
has 
squares, so Figure 100 had 
squares.
,
.
,
.