and 
Prove that
is divisibly by 
and 
be two infinite sequences of integers. Prove that there exists an infinite sequence of integers 
such that for any positive integer 
,![\[
\sum_{k|n} k \cdot z_k^{\frac{n}{k}} = \left( \sum_{k|n} k \cdot x_k^{\frac{n}{k}} \right) \cdot \left( \sum_{k|n} k \cdot y_k^{\frac{n}{k}} \right).
\]](http://latex.artofproblemsolving.com/2/5/9/259ee32c89befe7a3ee2797e0d5b1df1c3066cf6.png)
such that for all reals 
Proposed by Aritra12, India
and 
imply that either 
or 
.
implies that 
.
also implies that 
implies that for any 
,
or 
.
,
is injective. 
and 
imply that 
for all 
.
.
such that 
,
gives 
for all such 
.
then gives 
and thus 
.
which means that 
.
for all 
for some 
,
which is impossible. Thus 
in this case.
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to 
are collinear, and (iii) 
.
bisects 
.
for Girls, PRISMS Online Math Meet for Girls, is happening this spring! 
,
,
that does not contain 
,
and 
denote the incenters of triangles 
and 
,
passes through a fixed point.
Y by jhu08, HWenslawski, megarnie, Adventure10, GeoKing, QueenArwen, Rounak_iitr
Let 
be a positive integer. On the table, we have 
ornaments in different colours, not necessarily of each colour. Prove that we can hang the ornaments on Christmas trees in such a way that there are exactly ornaments on each tree and the ornaments on every tree are of at most 
different colours.
be a positive integer. On the table, we have 
ornaments in different colours, not necessarily of each colour. Prove that we can hang the ornaments on Christmas trees in such a way that there are exactly ornaments on each tree and the ornaments on every tree are of at most 
different colours.This post has been edited 1 time. Last edited by Morskow, Dec 17, 2017, 10:35 PM
of 
positive integers with sum 
and returns a set of 
positive integers with sum 
(call it an 
and 
respectively). Delete 
,
.
.
be the 
be the number of ornaments of the corresponding color. Repeatedly applying the "n-flow" to 
will eventually result in the empty set; it is not hard to see how this leads to the desired result.
be a positive integer. On the table, we have 
ornaments in 
different colours, not necessarily 
,
.
,
and 
,
,
.
ornaments of the maximum color to the first christmas tree. Now, we only have 
ornaments with 
colors (notice that one colour was gone) and we want to put these to 
Christmas trees in such a way that there are exactly 
)
marbles each of which is one of 
colors. We will prove that they may be placed in 
this is trivial.
then this implies the case for 
rectangle, we represent each box as a column of this rectangle, thus the task translates into filling this 
colors are used, then we just put any other color in the 
Thus at least one number of one colored marbles must exist that is 
Then you just pick that color to tile the 
and then we can just tile the remaining 
and win.
,
.
.
.
then each color has n balls and putting all balls of a color in the same box solves the problem. Thus, assume the contrary.
and 
.
balls of color n in the first box.
.
,
,
and let there be 
ornaments of the 
th colour. Assume wlog that 
.
such that 
,
and 
to form a group so that we use all of the ornaments of 
)
ornaments, so by the induction hypothesis, we can partition the rest.
for all 
,
.
.
colors has exactly 
A color 
which has less than 
marbles colored in it;
which has more than 
marbles colored in it.
marbles in color 
marbles in 
as well, which finishes. 
marbles in color 
marbles colored 
pretty much anything works since there are only 
marbles.
and the color with maximum marbles 
marbles from the maximum color (note that 
marbles of 
colors (since the minimum color one is gone) to be put in 
in our generalized problem we get the required result.
.
.
with more than or equal to 
as the number of marbles of color 
.
,
.
ornaments of color 
remaining colors, while there are 
remaining marbles. This means that there exists at least one color occupying at most 
balls of color 
and suppose 
.
, fill up the rest of 
.
.
.
.
.
'
marbles of at most 
is trivial, as you only have one color, so there's only one way to sort (and hence this way of sorting exists). Now let's suppose that you can sort as desired for 
;
.
marbles of that color (red), and another that has 
marbles of that color (blue). Then take all the red marbles, put them into a bin, and if there aren't 
ornaments, 
.
.
ornaments 
there is at least one colour (call it a big colour) of which there are 
ornaments . Take a small colour and put all of its ornaments on a tree. Fill the rest of the tree with ornaments of a big colour. Now we have used up 
ornaments, 
rectangle, so that we have 
,
.
and 
is forced. We fill an unfilled box with 
.
.
be an integer. There are 
and at most 
.
,
Let our inductive assumption be that it is possible for 
Now, for 
if the most abundant color has 
marbles and the least abundant has 
letting the last box have 
we obtain the desired.
is trivial.
if 
holds. By Pigeonhole, there exists one color 
ornaments of 
ornaments, where each color has strictly greater than, or less than 
ornaments, and 
be the number of marbles of each color. We use the following algorithm: While 
marbles of second color), for this we need 
,
marbles. Proof for 
.
the color corresponding to 
done! Also, at each step our algorithm decreases the sum by 
.
.
,
.
marbles of that color, and similarly there is a color so that there are 
marbles of that color. We can take all of the marbles of the color that has 
marbles of color 
marbles of color 
.
marbles of color 
and 
.
marbles, each of which is one of 
colors (color 
