AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
with 
, and announces them to the deputy leader and a contestant. The leader then secretly tells the deputy leader an -digit binary string, and the deputy leader writes down all -digit binary strings which differ from the leader’s in exactly positions. (For example, if 
and 
, and if the leader chooses 
, the deputy leader would write down 
and 
.) The contestant is allowed to look at the strings written by the deputy leader and guess the leader’s string. What is the minimum number of guesses (in terms of and ) needed to guarantee the correct answer?
be an acute scalene triangle. A point 
varies on its side 
. The points 
and 
are the midpoints of the arcs 
and 
(not containing ) of the circumcircles of triangles 
and 
, respectively. Prove that the circumcircle of triangle 
passes through a fixed point, independent of the choice of on .
are positive integer.
touches the sides 
, 
and at 
, 
and 
respectively. Let 
denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
with 
has at least two cycles in it.
that satisfy the following conditions:
is an interval.
be an acute scalene triangle inscribed in a circle 
. The angle bisector of 
intersects at 
and at 
. The point 
is the midpoint of 
. Let 
be the altitude in 
, and the circumcircle of 
intersects again at . Let 
be the midpoint of , and let 
be the reflection of with respect to . Prove that the triangles 
and 
are similar.
such that 
divides 
for any positive integers 
and 
.
for which the sequence of real numbers with 
for all 
is periodic from the beginning.
be an acute triangle and 
is orthocenter. Let be the intersection of 
and 
and be the intersection of 
and . The circumcircle of 
cuts the circumcircle of at 
. Prove that the angle bisectors of 
and 
concur at a point on 
is the family of all -tuples 
, where 
is a subset of a set containing 
elements. We can easily find that 
.
is the family of all -tuples , where is a subset of a set containing 
elements. We also have 
.
, the number of -tuples , where is a subset of 
, and with 
is 
.
.
, and let be the family of all -tuples 
such that each 
is a subset of 
. Let 
denote the number of elements of the set 
. Find![\[ \sum_{(A_1, A_2, \ldots, A_n)\in \mathcal{F}} |A_1\cup A_2\cup \cdots \cup A_n| .\]](http://latex.artofproblemsolving.com/c/5/e/c5e378912b3776a8254adb975059c41fab35525a.png)
subsets 
with 
, 
. For each such , the number of elements 
with 
is obviously 
. Therefore the sum we seek is![\[ \sum_{k=1}^N \binom {N} {k} k(2^{n} - 1)^k = N(2^n-1)\sum_{k=1}^{N} \binom {N-1} {k-1} (2^{n} - 1)^{k-1} = N(2^n-1)2^{n(N-1)} .\]](http://latex.artofproblemsolving.com/e/8/3/e83268a877d3c85c225547adb615e670a16a758c.png)
. The coefficient of 
is the number of -tuples 
of subsets of 
such that 
has elements and that 
. We want to associate 
with , and this can be done via differentiating 
with respect to 
and then replacing by 
. The sum of 
is then ![\[\left.\frac{\partial}{\partial t} F_{m,n}(t,1) \right|_{t:=1} = m \left(2^n-1\right)2^{(m-1)n}\,.\]](http://latex.artofproblemsolving.com/a/a/d/aadf3c098f2dd36ed9518a1e01e7695bfbdf9d9d.png)
, then one can differentiate 
with respect to each of its variables: 
And this is exactly the value of the sum.
be the set of all -tuples 
such that each is a subset of 
. Let denote the number of elements of the set . Find![\[ \sum_{(A_1, \ldots, A_n)\in F} |A_1\cup A_2\cup \cdots \cup A_n| \]](http://latex.artofproblemsolving.com/a/0/8/a0802b66ee9ec20913bbbf2c4a2ba546ff0e0dd3.png)
, the probability that 
is clearly 
. Thus the expected cardinality of 
is 
. The summation evaluates to
be the indicator of the proposition , i.e. 
and 
iff is true. Set 
and observe that
from symmetry. Now, there are 
-tuples 
. Of all those, 
do not contain . So, the number of those -tuples containing is![\[
2^{Nn}-2^{(N-1)n} = 2^{(N-1)n}(2^n-1),
\]](http://latex.artofproblemsolving.com/f/c/3/fc3075cbc7cb2a23d9e52ad65a00d01d0931add6.png)
yielding the desired answer as 
.
different subsets. Thus, there are 
different -tuples of .
. The probability that does not appear in is 
. The probability is doesn't appear in is 
. Therefore, the probability is appears in is 
. Hence, the desired sum is equivalent to
is 
There are numbers, so on average each summand produces 
Hence, as there are tuples in total, the answer is 
is a pair of real numbers such that
.
chance it is not in 
,![\[| A_1 \cup A_2 \cup \ldots \cup A_n | = 1998 \left(1 - \frac{1}{2^n}\right).\]](http://latex.artofproblemsolving.com/b/3/8/b3871f011b0e830616bbb13d8c5e08f5da1a9ffd.png)
.
where 
is an indicator function which determines whether 
is true or not. Now, notice that, by the Linearity of Expectation, we need to find the sum over all integers from 
Notice that there is a 
probability that 
so there is a 
probability that this indicator function is 
Thus, the expected value is
giving our answer of 
,
,![\[X_{i, A} = \begin{cases}1 & \text{if }i \in A_1 \cup A_2 \cup \dots \cup A_n \\ 0 & \text{otherwise}\end{cases}\]](http://latex.artofproblemsolving.com/c/6/6/c66ac69052e1fe4cd64dda2f6e4450d6a2babad2.png)
for each 
.![\[\sum_{A \in F} |A_1 \cup A_2 \cup \dots \cup A_n| = \sum_{A \in F} \sum_{i = 1}^n X_{i, A} = \sum_{i = 1}^n \sum_{A \in F} X_{i, A}.\]](http://latex.artofproblemsolving.com/7/f/e/7fe75b3304f19e95a8410e0556109d8523378cc4.png)
But for a fixed 
,
counts the number of tuples 
for any 
,
.![\[\sum_{A \in F} X_{i, A} = 2^{1998n} - 2^{1997n}\]](http://latex.artofproblemsolving.com/4/6/8/4682d5c1ce787e5f9d94790d6624cbb75b8f6003.png)
and thus that our desired sum is