AoPS Academy
Let ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
denote the set of natural numbers less than or equal to 
![$f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$](http://latex.artofproblemsolving.com/f/d/f/fdf7b2fcfb960e93d59f3f84cef4372262c69499.png)
there exists an -digit number divisible by 
all of whose digits are odd.
be an acute triangle with 
. Let 
be the centroid of triangle 
and let 
be the foot of the perpendicular from 
to side 
. The median 
intersects the circumcircle 
of triangle at a second point 
. Let 
be the point where intersects . The line 
intersects the circle at a point 
, such that lies between and . Prove that the points 
and lie on a circle. so that both and 
have odd numbers of divisors.
which is defined as all the natural numbers listed in order, side by side, behind a decimal point, without commas? If so, what is it?
and 
are primes, then can 
be a perfect square?
be a trapezoid such that 
lie on a circle with center 
, and side 
is parallel to side 
. Diagonals 
and 
intersect at point 
, and 
. It is given that 
. It is also known that the difference in length between and can be expressed in the form 
, where 
and 
are positive integers and is square-free. Compute the value of 
.
, where 
always occurs at 
, above the xy-plane, such that the pyramid 
is a regular tetrahedron.
and are searching for 
. We can not take only of form
.
forbidden values of , which correspond to 
, and 
forbidden values which correspond to 
. So, at most 
forbidden values, and at least 
admissible., you always take the smallest that is not forbidden, you will only have to make sure that is distinct from all 
. This is true because the values with 
are all 
, and so have been forbidden at some earlier step.
.
such that the sets are parwise disjoint even when taken 
.
.
iff the sets 
and 
are disjoint . The graph will have 
vertices . For an arbitrary vertex 
to be joined with 
, 
must not be one of the numbers 
, which are 
numbers . So the vertex has degree at least 
.
edges. The problem asks to prove that there is an 
in the graph. But by Turan's theorem , there is a 
in a graph with vertices iff the number of edges is strictly greater than :
to be the remainder of when divided by 
.
. A simple calculation shows that the number of vertices is greater than 
and thus we are done.
such that 
doesn't contain 
and any element 
satisfying 
for any distinct 
, but the case remains where 
, which can also satisfy 
instead of 
. So the limit is (about) doubled and it reaches near , which can be achieved with a little more strict bounding.
such that doesn't contain and any element satisfying for any distinct , but the case remains where , which can also satisfy instead of . So the limit is (about) doubled and it reaches near , which can be achieved with a little more strict bounding. is the smallest element in 
and any other element in is greater than .So, there is no satisfying
's in increasing sequence, then it reduces to 
where 
which implies 
and thus the strategy is optimized by two by the algorithm, for the sake of contradiction , let's assume there exists 
and 
such that 
. when we choose 
(According to algorithm).
". Thanks for your concern.
be the vertex of a graph where two vertices 
are connected by an edge if and only if the sets 
and 
are disjoint. Consider an arbitrary vertex 
. Since the 
the maximum number of vertices 
such that sets and 
are not disjoint is 
. Therefore every vertex of the graph has degree at least 
Therefore the graph has at least 
edges. It suffices to show that this graph contains 
as a subgraph.
vertices that does not contain may contain is obtained when the graph is a complete 
-partite graph with 
independent sets of size 
and 
independent set of size 
. It is easy to compute that this graph has 
edges. But since 
we have the desired result.
numebr on his list? Write its last two digits as your answer.
grid of square tiles. Robert chooses four of these squares uniformly at random. The probability that the centers of these four squares form a square themselves is 
where 
are coprime naturals. Write the largest 
digit odd factor of 
.
.
vertices of the face opposite to it. Find the integer closest to 
are distinct numbers in AP such that 
are also in AP. Find the smallest possible value of 
(in degrees).
that satisfy 
be a 
-
.
,
in ![\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \]](http://latex.artofproblemsolving.com/d/5/c/d5c2a7b943e8c7200361698f3eb6f83700ad68d6.png)
are pairwise disjoint.
.
and 
values of 
with distinct 
such that 
doesn't contain 1 and any element 
for any distinct 
Choose the smallest element from 
.
satisfy the condition and the interesting fact is for choosing all 
.
.
.
be the first value such that 
for all ![$j \in [1, n]$](http://latex.artofproblemsolving.com/9/7/c/97c9a519a3453c51bdd4804d10c37a3c9154e90c.png)
.
,
.
is bounded above by 
.
for all valid 
;
for all valid 
and 
.
distinct absolute differences between two elements of 
with common difference 
,
.
are 
;
,
)
.
be the numbers, and let 
be the numbers. Let 
,
.
since if 
we have at least one more valid choice to add to 
,
satisfying the problem conditions for maximal 
and 
for some 
or 
for some integers 
and 
.
,
ways to choose 
,
,
.
such that, for any choice of 
from 
and 
)
and 
have at least one common element.
be the sets 
respectively. For each 
can be expressed as 
.
:![\[ B = B_1 \cup B_2 \cup \ldots \cup B_{100} \]](http://latex.artofproblemsolving.com/9/3/e/93e422a9a40e45dff563a69a86b7329350ce9600.png)
,
'
.
,
.
.
,
,
,
we have 
.
(
as desired.
we cannot pick 
,
in the list of 
different 
distinct absolute differences in 
elements get eliminated from 
,
.
,
numbers and see that we have eliminated 
numbers, and there is one number remaining and we see that we can just add this and done.
.![\[t_i-t_j \notin \{a_p-a_q\vert 1\le p \ne q \le 101\}.\]](http://latex.artofproblemsolving.com/0/a/4/0a472d7d6821c451015fb7852db17ae9b0e2d787.png)
such that 
is as large as possible. That means every 
is either in ![\[i = t_i+b-a,\]](http://latex.artofproblemsolving.com/8/8/a/88a20f9ef2f6f0ef8339d90f6a38bc1f03ebee14.png)
and 
.
values of ![\[n+10100n \ge 10^6 \implies n>99,\]](http://latex.artofproblemsolving.com/5/a/7/5a77a3000be63f5d47a116472456b61b4186404e.png)
currently. Let 
.
denote the pair-wise differences of elements of set 
for some 
.
.
and we are done.
and let the elements of 
Since each 
is distinct, then clearly for all 
Set 
Clearly since there are at most 
every time we add a number to the set 
(we account for both sides, and the number itself) Therefore, after choosing 
possible number to add to the set 
so thus it is possible to have 
QED
.
satisfying the condition, then we can choose a 
and 
However, since 
,
.
ways to choose the right-hand side of the equation 
.
,
numbers that 
so there always exists a 
,
are max such subsets.
has an element common with one of 
,
,
.
,![\[a_\ell+T=a_j+t_i\]](http://latex.artofproblemsolving.com/3/d/f/3dfcaa495584457f85fbd7f968b8401a36c203c0.png)
Fix RHS and see that 
takes 
values and so ![\[101k \ge \frac{10^6-k}{100} \implies k>99 \iff k \ge 100\]](http://latex.artofproblemsolving.com/3/a/4/3a41c5d60546635d95be6306d2d8235d1dee37c5.png)
And done.
be the elements which can be expressed as the difference of two elements in the list. Then 
since 
is over counted 
for the sets to be pairwise disjoint, and furthermore this is sufficient. As such, we may select 
while preserving the condition. For each 
so we may choose