table, several cells are marked. At each move, Cyril can get to know the number of marked cells in any checkered square inside the initial table, with side less than 
.
be triangle with circumcenter 
and orthocenter 
.
intersect 
respectively. Lines through 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
be an odd integer. Prove that there exists a prime number 
such that![\[
p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n.
\]](http://latex.artofproblemsolving.com/3/5/1/3514b64ecd5f7b2586d5bfeae59b7c9504d5b455.png)
and 
with equal radii intersect at P and Q. Points B and C are located on the circle
The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
![\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]](http://latex.artofproblemsolving.com/2/d/2/2d248ce9756f0f09265dd538de3b16b22a944c46.png)
![\[ \begin{tabular}{c c c c}
{} & \textbf{Adams} & \textbf{Baker} & \textbf{Adams and Baker} \\
\textbf{Boys:} & 71 & 81 & 79 \\
\textbf{Girls:} & 76 & 90 & ? \\
\textbf{Boys and Girls:} & 74 & 84 & \\
\end{tabular}
\]](http://latex.artofproblemsolving.com/8/5/c/85c75770c3f90af49df0db600dead2a29e6674b3.png)
Y by
Problem Statement:
The set S= {1/r : r = 1, 2, 3,...} of reciprocals of the positive integers contains arithmetic progressions of various lengths. For instance, 1/20, 1/8, 1/5 is such a progression, of length 3 (and common difference 3/40). Moreover, this is a maximal progression in S of length 3 since it cannot be extended to the left or right within S (−1/40 and 11/40 not
being members of S).
(i) Find a maximal progression in
S of length 1996.
(ii) Is there a maximal progression in
S of length 1997?
For part (i) I constructed the sequence: 1/1996!, 2/1996!, 3/1996!, ..., 1996/1996!. Which has common difference of 1/1996! and because of the nature of the factorial function, each of them are elements in S, and continuing to the left we get 0, which is not in S, and continuing to the right we get 1997/1996! which is not in S because 1997 is prime. Therefore we have found a maximal progression in S of length 1996 as desired.
For part (ii), let A = (4000!)/(2003!). Now consider the sequence: 2004/A, 2005/A, 2006/A, ... , 4000/A. Each is an element of S, by the nature of the factorial function and the common difference is 1/A. Continuing to the left gives 2003/A, 2003 is prime which is not a factor of A, so 2003/A is not an element of S. Continuing to the right gives 4001/A, 4001 is prime which is not a factor of A, so 4001/A is not an element of S. Therefore we have found a maximal progression in S of length 1997, and so the answer is Yes.
Is this proof sound?
The set S= {1/r : r = 1, 2, 3,...} of reciprocals of the positive integers contains arithmetic progressions of various lengths. For instance, 1/20, 1/8, 1/5 is such a progression, of length 3 (and common difference 3/40). Moreover, this is a maximal progression in S of length 3 since it cannot be extended to the left or right within S (−1/40 and 11/40 not
being members of S).
(i) Find a maximal progression in
S of length 1996.
(ii) Is there a maximal progression in
S of length 1997?
For part (i) I constructed the sequence: 1/1996!, 2/1996!, 3/1996!, ..., 1996/1996!. Which has common difference of 1/1996! and because of the nature of the factorial function, each of them are elements in S, and continuing to the left we get 0, which is not in S, and continuing to the right we get 1997/1996! which is not in S because 1997 is prime. Therefore we have found a maximal progression in S of length 1996 as desired.
For part (ii), let A = (4000!)/(2003!). Now consider the sequence: 2004/A, 2005/A, 2006/A, ... , 4000/A. Each is an element of S, by the nature of the factorial function and the common difference is 1/A. Continuing to the left gives 2003/A, 2003 is prime which is not a factor of A, so 2003/A is not an element of S. Continuing to the right gives 4001/A, 4001 is prime which is not a factor of A, so 4001/A is not an element of S. Therefore we have found a maximal progression in S of length 1997, and so the answer is Yes.
Is this proof sound?
