such that 
divides 
for all positive integers 
and 
with 
.
be a triangle with circumcenter 
and internal bisector 
.
again at 
and 
cuts 
.
lie on 
.
at 
respectively. Let circumcircle of 
cuts 
.
.
such that the number
is the product of two or more consecutive numbers.
and 
at the points 
and 
,
be the point where the lines 
and 
meet, and let 
and 
be points such that the two quadrilaterals 
and 
are parallelogram.
.
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Determine all real numbers 
such that, for every positive integer 
the integer
is a multiple of 
(Note that 
denotes the greatest integer less than or equal to 
For example, 
and 
)
Proposed by Santiago Rodríguez, Colombia
such that, for every positive integer 
the integer
is a multiple of 
(Note that 
denotes the greatest integer less than or equal to 
For example, 
and 
)Proposed by Santiago Rodríguez, Colombia
This post has been edited 2 times. Last edited by EthanWYX2009, Jul 19, 2024, 5:33 AM
Reason: change to original text
Reason: change to original text
.
,
.
Then 
is integer and so 
.
and so 
.
and get 
for all 
.
and thus 
for all 
.
for large 
then 
is divisible by 
also satisfies hence we can assume that 
.
.
denote 
as 
where 
.
But 
which is impossible. Also 
doesn't hold which can be seen by taking 
.
.
where 
.
Let 
We have 
since 
yields 
.
.
.
for all 
then 
for 
.![\[\frac{m+1}{2}|m+1|\frac{(m+1)(m+2)}{2}+\frac{(m-1)m}{2}+\lfloor (m+1)\beta \rfloor\equiv \frac{m^2-m+m^2+3m+2}{2}+\lfloor (m+1)\beta \rfloor\equiv 1+\lfloor (m+1)\beta \rfloor\]](http://latex.artofproblemsolving.com/9/2/4/9243ed7fea2de389764ba54f9ae2aea56862db88.png)
Since 
and 
we get 
for all 
where 
as desired
As i was thinking about this in my head, I just used the stupidest bound that I could think of, namely 
to get that 
.
belonged to the interval ![$\left(\alpha\frac{n(n+1)}{2}-n,\alpha\frac{n(n+1)}{2} \right]$](http://latex.artofproblemsolving.com/4/a/6/4a6311ff450930ad03d360f9fc5fdb372a2c9660.png)
which has length 
in the easiest way possible wouldn't work because the orders of magnitude didn't match. At this point, I arrived to the gym, so I didn't think more about the problem, but thought it had potential to be a nice problem.
.
.
is not a multiple of 2.
.
be the smallest integer such that 
;
.
,
,
.
;
.
,
where 
and 
.
from here we see clearly that 
for all 
therefore 
.
and 
a contradiction as we know 
on RHS is linear function of higher slope.
only, for 
.
.
.
for some integer 
,
.
Therefore, 
is congruent to 
modulo 
for all 
.
for all 
.
which is clear because 
have 
.
is divisible by 
.
,
.
must be divisible by 
,
for all 
for all 
for all 
i.e. 
,
for some integer 
.
(we will verify them later).
.
for some integer 
.
so it follows that 
for all 
,
.
is divisible by 
,
.
,
Since 
,
,
for all 
,
,
contradicting the fact that 
which is indeed divisible by 
.
.
,
.
.
but 
.
,
.
,
,
,
so this case also doesn't work out.
.
. Thus,
,
. So,
. We can express 
,
.
.
, so 
.
,
for all k, which means 
.
for any 
.
in 
can be computed. In order for there to be more solutions other than 
should be divisible by n 
.
is also one. Therefore, the claim is true. 
![\[
\lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \cdots + \lfloor n\alpha \rfloor
\]](http://latex.artofproblemsolving.com/2/9/7/29763bad88095eb3d2992daaaf20502743b68915.png)
ifoda 
orqali 
dan kichik yoki teng bo’lgan eng katta butun sonni belgilaymiz. Masalan, 
va 
.
for any integer ![$$[\alpha+2k]+...+[n\cdot \alpha+2nk]=[\alpha]+...+[n\cdot \alpha]+k(2+...+2n)$$](http://latex.artofproblemsolving.com/6/6/8/668c5027c2fba4ed55fdc7a901e4bfd85bd46ff2.png)
where 
.![$\alpha \in [-1,1]$](http://latex.artofproblemsolving.com/1/a/9/1a90a60816adc5ed0704a98cf5b9f303d16f9f7b.png)
before generalizing for all reals.
works but 
don't when 
such that 
.![$$[\alpha]+...+[n\cdot \alpha]=0+...+0+1=1$$](http://latex.artofproblemsolving.com/b/6/d/b6dc4670a2524184de12adacdca708e93531b744.png)
which is not divisible by 
.![$$[\alpha]+...+[n\cdot \alpha]=(-1)+...+(-1)+(-2)=-(n+1)$$](http://latex.artofproblemsolving.com/8/0/7/8071e651bf503fd5c5fb718bd16b7905bec0b34c.png)
which is not divisible by 
so 
must be an integer. Since 
.
if 
.
is a multiple of 
,
.
Since this is a multiple of 
,
,
works it implies 
and assume it works, then we have 
If we observe 
and so forth it will imply for all 
,
Now let 
then we have 
, if 
then we have 
.
for some 
.
.
odd, this only holds true if 
.
.
.
,
.
and 
.
,
and continuing this we have 
,
and consider 
.
,
.
.
So 
.
and 
has to be even for 
,
.
and in fact 
,
,
or 
,
are easy to check. Suppose it holds true for 
.
As 
,
However, 
and 
,
.
,
.
,
,
.![$[0, 2]$](http://latex.artofproblemsolving.com/a/3/b/a3b958703526b6271d0f6ac63af4fec05de1acdf.png)
because for anything larger we can take out a 
from the sum of floors which dies mod ![$[0, \frac{1}{n}) \cup [2 - \frac{1}{n}, 2]$](http://latex.artofproblemsolving.com/d/d/f/ddfe4cf31a501c7dc9e0d49680f384fd4a011356.png)
.
because of trivial base case. Notice that the good interval contains 
because the sum is zero. Now we need to prove that everything in 
fails because the other side is basically the same reasoning. This is true because the sum here has the 2nd to last term in the sum equal to zero still but the last term is not enough to push it to a multiple of 
also does. Hence WLOG assume that 
It is easy to see that 
does not work by setting 
If 
it clearly works.
let 
Then setting 
yields 
impossible. Similarly 
does not work too.
works so the answer is all even integers 
for 
,
has integer part 
.
to denote the given sum truncated at 
be an odd integer. Then 
divides 
if and only if it also divides 
.
,
.
,
for all 
.
,
as only the 
and 
,
cannot satisfy the conditions.
.
,
for all 
too. It follows that ![\[S_{2n+1}(r) = 1 + 2 + \cdots + 2n-2+ 2n-1 + 2n-1 - \varepsilon = n(2n+1) - 1 - \varepsilon\]](http://latex.artofproblemsolving.com/2/7/6/276a56314116a53c16b0c00d5c6e72b4824d0919.png)
for some 
depending on the value of 
.
,
