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be a quadrilateral inscribed in a circle 
. Let 
, 
, 
, 
, 
, 
, 
, 
. Prove that 
lies over 
.
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist. be an odd prime and 
be integers so that the integers 
are divisible by . divides each of .
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
are in arithmetic progression. Find the value of 
.
such that![\[
\frac{11}{\log_2 x} + \frac{1}{2 \log_{25} x} - \frac{3}{\log_8 x} = \frac{1}{\log_b x} \quad \text{for all } x > 1.
\]](http://latex.artofproblemsolving.com/2/d/7/2d76319b3d4fcd1f193ea0f9f37cb4bf513376f3.png)
![\[ x^{\log{x}} = 10^{2 - 3\log{x} + 2(\log{x})^2}\]](http://latex.artofproblemsolving.com/b/f/b/bfb8185992867a8926b5c62198a799c935771d53.png)
where 
is the logarithm of 
to the base 
that satisfies![\[
a^3+b^3=1957
\]](http://latex.artofproblemsolving.com/5/d/8/5d8d2a40734ec6b0a9149f9a4ae3a5828d5055f3.png)
![\[
(a+b)(a+1)(b+1)=2014
\]](http://latex.artofproblemsolving.com/3/0/0/300f2d16e669a1d8d9c54b357a6eaf7a8eb5d826.png)
Hence, find the value of 
students. Let, for 
, the 
student have 
many friends. For 
let us define to be the number of students who have strictly more than friends. Show that 
for all , but 
(and all other 
), so the equation becomes 
. Perhaps means the number of students with strictly more than friends. is counted once for all 
, namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
friends is counted in 
.i.e times . so in the RHS too it is counted times . so these are two different ways of counting the sum 
and hence it is proved
chaotic_iak already gave that "simple solution" of yours. Why bother repeating it?
for all , but (and all other ), so the equation becomes . Perhaps means the number of students with strictly more than friends. is counted once for all , namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
-gon with vertices 
(
) as persons.We colour the segment 
red if and 
are friends,otherwise we colour it as blue.We shall now show that both the sides number twice the number of red sides..It is counted once when is counted and once when 
is counted.Thus LHS part is proved.
.While we are counting 
,we are only counting 
redlines emanating from 
.This proves our claim for RHS.(Note that thus summing over the 's gives the sum of exact number of redlines originating from each vertex).
means the number of students with strictly more than friends. , then what is 
? 0 ?
be a simple graph with 
vertex. Suppose 
And we connect 
if they are friend 
. Suppose 
.And 
denote number of vertices having degree more than 
.
.
.
.
.
.If it is isolated then 
,
count number of vertex having degree more than in the new graph with vertex. we are done. is not isolated. . Suppose it joins with 
,
. Then 
.
,
.
will as same as .
.One must see that number of edges in new graph is 
.
new edges
.
..
for all , but (and all other ), so the equation becomes . Perhaps means the number of students with strictly more than friends. is counted once for all , namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
is NOT equal to , it is indeed 
. Observe the question carefully.. means the number of students having MORE than (i.e. 
) friends.. which according to your assumption that everyone hates everyone is indeed . For the same reason, the iteration of stops at 
and not because no one can possibly have 
friends.. hope I could make myself clear..
students. Let, for , the student have many friends. For let us define to be the number of students who have strictly more than friends. Show that 
means the number of students with strictly more than friends. , then what is ? 0 ?
and 
.
Prove that
This is unexpected. And sad. Because I just proved a wrong problem. 
array with the columns named as 
and the rows named as 
.
.. Suddenly the problem becomes so much more interesting, isn't it ? 
.
,
boxes, i.e. all the boxes below 30. So to the count we add 30, and to the boxes, we're adding 30 balls.
.
number of students who have greater than or equal to 
)
if 
or,
if 
