[/quote]
,
is called a 
digits in decimal representation (it cannot have leading zeros).
is a 
-![\[(20 + 25)^2 = 2025.\]](http://latex.artofproblemsolving.com/5/c/5/5c51bca485cea6c3750c48c09091eb5fd72b5192.png)
Find all 
-
be the set of all positive integers greater than or equal to 
and let 
be a function such that 
for all 
if 
,
players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least 
next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
be a fixed positive real number. Find the minimum possible value 
such that for any positive reals 
,
,
,
,
and 
divides 
.
and 
.
such that 
.
,
,
,
.
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For 
represents 
.
![\[
\frac{5^{2000} - 1}{4}
\]](http://latex.artofproblemsolving.com/8/b/a/8ba5d1ed8bafd6653f9e6f351558a9976c43a57a.png)
is divided by 
.
be the circumcircle of acute triangle 
.
and 
are on segments 
and 
respectively such that 
.
and 
intersect minor arcs 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.
is a positive real number, find the minimum of the following expression
Y by Wave-Particle, mathmaths, rightways, baladin, WhaleVomit, Davi-8191, rterte, kk108, Tawan, Mathuzb, megarnie, Rounak_iitr, NO_SQUARES, Adventure10, Mango247, Sedro
The equation
is written on the board, with 
linear factors on each side. What is the least possible value of 
for which it is possible to erase exactly of these 
linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
is written on the board, with 
linear factors on each side. What is the least possible value of 
for which it is possible to erase exactly of these 
linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?This post has been edited 2 times. Last edited by sseraj, Jul 12, 2016, 5:44 PM
If one erases all the 
in LHS and 
in RHS, then it can be proved that the remaining equation doesn't have real roots.
.
in 
where 
,
.
.
,
。
,![\[\prod_{i=-p}^{q}f(a-4i)=1\]](http://latex.artofproblemsolving.com/1/5/2/152bfa0d35a607fe6eb8bf2e3f0fe8c2b409b1e9.png)
where 
are non-negative integers (I'm too lazy to write down its exactly form).![\[\prod_{i=-p}^{-1} f(a-4i)=\prod_{i=1}^{p} f(a+4i) = \prod_{i=1}^{p} (1-\frac{2}{(a+4i)^2-0.25})\geq 1-\sum_{i=1}^{p} \frac{2}{(a+4i)^2-0.25}\]](http://latex.artofproblemsolving.com/e/c/2/ec22e739ab8c81b98a1058ed73bae7629adb02a0.png)
![\[>1-\frac{1}{6}-\sum_{i=2}^{\infty}\frac{2}{(a+4i)^2-0.25}>1-\frac{1}{6}-\frac{1}{8}\sum_{i=1}^{\infty}i^{-2}=\frac{5}{6}-\frac{\pi ^2}{48}>\frac{1}{2}\]](http://latex.artofproblemsolving.com/6/9/8/698794ee8808704f73d76c3a6e8c9fbef3e17044.png)
Similarly, 
.
,
,
,
,
.
,
.
inequalities
Notice that in all these inequalities, at most one of them has non-positive numbers in it, and we never have both zero. So it is OK to multiply them all together, and we get the conclusion. Thus we have a construction for 
,
the LHS and RHS must have two coinciding linear terms.
)
sides, it is on the left and we can multiply all together. Thus the only case that remains is if 
for some 
.![\[ \prod_{k \ge 0} \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)} < e. \]](http://latex.artofproblemsolving.com/b/5/4/b54646631ad989e48589233f6ae72f311a4f2455.png)
To see this, note that it's equivalent to prove ![\[ \sum_{k \ge 0} \log \left( 1 + \frac{2}{(4k+1)(4k+4)} \right) < 1. \]](http://latex.artofproblemsolving.com/a/4/2/a4240564e98d58606835805a5eb1399f3ea7d784.png)
To this end, we use the deep fact that 
,![\[ \sum_{k \ge 0} \frac{1}{(4k+1)(4k+4)} < \frac12 \]](http://latex.artofproblemsolving.com/c/d/a/cda961fe5889adda61bfee040f49515bfbf34eae.png)
which one can obtain for example by noticing it's less than 
.
,
since erasing any 
(or less) factors will leave a common linear factor on both sides.
have all the 
roots and 
the 
roots:![\[
\begin{array}{cccc}
p(x) := (x-1)(x-4)&(x-5)(x-8)\ \cdots \ (x-2013)(x-2016) \\
q(x) := (x-2)(x-3)&(x-6)(x-7)\ \cdots \ (x-2014)(x-2015).
\end{array}
\]](http://latex.artofproblemsolving.com/e/b/9/eb912e0cf57830e228ca0a8ac9bdd90f8a16afbd.png)
We need to check that 
has no real solutions. We claim 
for all 
.![\[
\begin{array}{c|cccc|cccc|c|c}
(-\infty,1) & (1,2)&(2,3) & (3,4) & (4,5)&(5,6)&(6,7)&(7,8)&(8,9)&\cdots &(2016,\infty) \\
&a & b & c & d & a & b & c &d &\cdots
\end{array}
\]](http://latex.artofproblemsolving.com/1/9/b/19b23e288462f8a2280bea54c242c78f0de2a8fb.png)
It suffices to use only open intervals since on the boundaries, exactly one of 
or 
.
and that for 
and 
,
.
interval of type 
,
.
The key is that for each row, both sides are positive. So multiplying gives 
,
and 
.
for some 
and 
for some 
,
construction, 
does not have degree 
,
for various 
.
doesn't work.
such that 
is on both sides of the equation after erasing, which implies that ![\[(x-1)(x-4)(x-5)(x-8)(x-9)(x-12)\cdots(x-2013)(x-2016)=(x-2)(x-3)(x-6)(x-7)(x-10)(x-11)\cdots (x-2014)(x-2015).\]](http://latex.artofproblemsolving.com/0/d/d/0dd5535f6f02beae034fa863a9f6f61d3f5698f9.png)
.![\[(x-1)(x-4)=x^2-5x+4<(x-2)(x-3)=x^2-5x+6, (x-5)(x-8)<(x-6)(x-7),\]](http://latex.artofproblemsolving.com/e/b/5/eb5efa237ba2fb09176d25b0b3d3d1130bd7b27b.png)
and so on until 
.
are all positive, which implies the right-hand side is strictly greater than the left-hand side.
.
don't have any solutions, where 
is a non-negative integer.
and 
,
don't have any solutions, where 
and 
)
don't have any solutions, where 
or 
)
don't have any solutions, where 
(we will put them back later). So the equation becomes ![\[(x-4)(x-5)(x-8)(x-9)(x-12)(x-13)\cdots(x-2012)(x-2013)=(x-3)(x-6)(x-7)(x-10)\cdots(x-2011)(x-2014).\]](http://latex.artofproblemsolving.com/b/4/2/b421ca8e9239ed0885374dcc3a5318147d90942d.png)
.![\[(x-4)(x-5)(x-8)(x-9)(x-12)(x-13)\cdots(x-2012)(x-2013)>(x-3)(x-6)(x-7)(x-10)\cdots(x-2011)(x-2014)>0.\]](http://latex.artofproblemsolving.com/6/5/0/65069459c656c30fa73bd02b3c2e36192d4969b8.png)
and 
,
.
(large positive number) = small negative number and RHS=(large negative number) 
remains on different sides of equality. We claim, that 
which suffices.
for 
By multiplying we are done.
is zero, hence the conclusion.![$x\in [1;2016]\backslash \mathbb{Z}.$](http://latex.artofproblemsolving.com/a/7/a/a7a1953b867fc01bad8f181a098804cd5532487e.png)
Here we consider two additional cases.
is odd.
we have 
Moreover 
so multiplying altogether we obtain 
will have no solution, there obviously must be no repeats on the 2 sides.
,
for the LHS and 
for the RHS.![$[1, 2, 3, \dotsm, 2016]$](http://latex.artofproblemsolving.com/1/f/2/1f2cdb78adc11d951aa91c376d78a11a86623d20.png)
,
![\[\prod_{k=1}^{504}{(x-(4k-3))(x-4k)}=\prod_{k=1}^{504}{(x-(4k-2))(x-(4k-1))}\]](http://latex.artofproblemsolving.com/3/9/6/396b43328a1baf91bef1adbac71439231fcc4eec.png)
has no real solutions. Clearly, there are no integer solutions between ![\[(x-(4k-3))(x-4k)=(x-(4k-2))(x-(4k-1))-2 < (x-(4k-2))(x-(4k-1)).\]](http://latex.artofproblemsolving.com/5/b/5/5b55107af582c6e0f3b885993bed40aa451ecc59.png)
Suppose 
then every factor on the LHS is positive and less than every factor on the RHS, so the LHS is less than the RHS. If 
for some integer 
then every factor on the LHS is positive except one, so it is negative. Every factor on the RHS is still positive, so it is positive. Again LHS 
RHS.
.
and 
are the only negative terms on each side. We have 
.
be from 
to 
then ![\[\frac{(x-(4(k-i)-2))(x-(4(k-i)-1))(x-(4(k+i)-2))(x-(4(k+i)-1))}{(x-(4(k-i)-3))(x-4(k-i))(x-(4(k+i)-3))(x-4(k+i))}=\frac{s^2}{s^2-4}\]](http://latex.artofproblemsolving.com/2/c/b/2cb5ddfeeaabbf768201cf4850e656433bee4a09.png)
and it is easy to see from here that no amount of this being multiplied together with distinct positive 
'
where 
and let the RHS contain the rest of the factors. Our equation is then ![\[\prod_{k=0}^{503}\left(x-(4k+1)\right)\left(x-(4k+4)\right)=\prod_{k=0}^{503}\left(x-(4k+2)\right)\left(x-(4k+3)\right)\]](http://latex.artofproblemsolving.com/2/e/6/2e6df08202e0e8f6147a10884f6c4ad1ae5d592e.png)
Notice ![\[\left(x-(4k+1)\right)\left(x-(4k+4)\right)+2=\left(x-(4k+2)\right)\left(x-(4k+3)\right)\quad(*)\]](http://latex.artofproblemsolving.com/7/b/0/7b03e0c83af240159f505ecfc3416fd2060c6ed0.png)
Hence,
then one side is zero and the other side nonzero.
for any 
then one side is negative and the other is positive.
,
for 
,
is a solution to the equation. Consider ![\[P=\prod_{0\le k\le 503,k\neq j}\frac{\left(x-(4k+1)\right)\left(x-(4k+4)\right)}{\left(x-(4k+2)\right)\left(x-(4k+3)\right)}=\frac{\left(x-(4j+2)\right)\left(x-(4j+3)\right)}{\left(x-(4j+1)\right)\left(x-(4j+4)\right)}=\frac{\epsilon^2-\epsilon}{\epsilon^2-\epsilon-2}<\frac{1}{9}\]](http://latex.artofproblemsolving.com/b/d/4/bd48586b33cbdf370df34c6d16b16a8920ed5e85.png)
Let 
be a term in 
.![\[f(j-\ell)\ge \frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\]](http://latex.artofproblemsolving.com/b/9/b/b9b9ea50496a571614af7f8de72422157f35a552.png)
and ![\[f(j+\ell)\ge \frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\]](http://latex.artofproblemsolving.com/6/a/d/6ad0f08a571c63666db5eb1ee6cf5b92b15ac091.png)
by 
.
increases as we get further from ![\begin{align*}
P&=[f(j-1)f(j-2)f(j-3)\dots f(j-1008)][f(j+1)f(j+2)\dots f(j+1008)]
\\&\ge\left(\prod_{\ell=1}^{1008}\frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\right)^2
\end{align*}](http://latex.artofproblemsolving.com/c/3/6/c360b36f104b15981c8d161e29246b8389721ecd.png)
It suffices to show ![\[\prod_{\ell=1}^{1008}\left(1+\frac{2}{(4\ell)(4\ell-1)-2}\right)<3\]](http://latex.artofproblemsolving.com/6/6/3/66338c417e0470bc501708aecf7892dbb97587af.png)
in order to obtain a contradiction. We know 
so
as desired. 
.
and RHS as 
.
(the key being for ANY x), so if both sides are negative LHS>RHS, if both sides are positive LHS<RHS, and both sides can't be zero, done.
,
,
.
must be erased, otherwise, 
,
from RHS and erase factors 
from LHS. Then we will prove that the equation 
has no real solution.![$x \not\in [1, 2016]$](http://latex.artofproblemsolving.com/f/b/f/fbf38a0b3dfffe78bbed89d46d1391cadc33b0b1.png)
,
,
.![$x \in [1, 2016]$](http://latex.artofproblemsolving.com/2/2/8/2281be6b459c825d1d38ca58871d44a6e40d4640.png)
.![$x \in [4s, 4s+1]$](http://latex.artofproblemsolving.com/b/b/2/bb2ff85969688d88f6e9064980e146a168bb11b4.png)
for some 
.
for all 
![$x \in [4s+1, 4s+2]$](http://latex.artofproblemsolving.com/9/4/c/94cdc8d442836a610221dbe804c57648a61eeb72.png)
or ![$x \in [4s+3, 4s+4]$](http://latex.artofproblemsolving.com/a/d/3/ad37e197226bc73d0110770c70c81c33aeb9766c.png)
for some 
and for all 
,
.![$x \in [4s+2, 4s+3]$](http://latex.artofproblemsolving.com/7/b/a/7ba958ecbedd4d183ab3e6c3381ad3d9323ba8ff.png)
for some 
,
.
from 2 sides and using the inequality 
,
.
,
.
,
.
We show that removing all with constant either 
from the left side and 
from the right side works. Note that for all 
we have that
We can simply multiply all the inequalities of this form together whenever 
However, when 
we have that for exactly one inequality the left and right sides are both negative, where the left is maximum the product of 
Note that 
Note that we have two of these products because 
so because 
we're don
we make the following construction:
If 
or 
then 
and all of these pairs of products are positive, so the left hand side is clear less than the right hand side. If 
but 
for some integer 
then the LHS is less than 
while the right hand side is not.
then
Now, we get that everything else is at most
Thus, this ratio cannot be 
We will prove that this has no solutions. Indeed
Suppose there is a root 
or 
,
or 
.
.
when compared, since all terms are positive, can be multiplied together
(Note that the actual product without inversion in signs is negative so the sign gets flipped)
appears on both sides.![\[P(x) = \prod_{i=0}^{503}(x-4i-1)(x-4i-4) - \prod_{i=0}^{503} (x-4i-2)(x-4i-3).\]](http://latex.artofproblemsolving.com/9/e/c/9ecdf5356715ea5b5d1ffebe66310607da10fe98.png)
We will prove that 
for all real numbers 
and 
denote the two products, such that 
.
when 
or 
for each 
.![$x \in [4k+1, 4k+2]$](http://latex.artofproblemsolving.com/c/7/d/c7df36c7ab9cdb6eb130fad2d53f0370c34c367b.png)
or ![$x \in [4k+3, 4k+4]$](http://latex.artofproblemsolving.com/c/3/3/c33e7408a8d5fd50b7f4abfe71d74767ad8df259.png)
for any integer 
,
is nonnegative and 
is nonpositive, with equality not holding simultaneously. When 
,
where each inequality follows again by multiplying the same inequalities in the claim.
.
and 
,
for these values of 
Hence we get the result.
,
on one side and 
on the other side works.
.
,
,
integer 
,
,
,
all succumb to both terms being positive or the inequalities 
,
.
and 
and let 
,
![$<\frac{1}{9}(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(4k-2)(4k+1)})(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(2010-4k)(2013-4k)})<1\longleftrightarrow \log [(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(4k-2)(4k+1)})(1+\frac{2}{2\cdot 5})(1+\frac{2}{6\cdot 9})\dots(1+\frac{2}{(2010-4k)(2013-4k)})]<\frac{2}{2\cdot 5}\frac{2}{6\cdot 9}\dots \frac{2}{(4k-2)(4k+1)}\frac{2}{2\cdot 5}\frac{2}{6\cdot 9}\dots \frac{2}{(2010-4k)(2013-4k)}=\frac{2}{3}(\frac{1}{2}-\frac{1}{5}+\dots +\frac{1}{4k-2}-\frac{1}{4k+1}+\frac{1}{2}-\frac{1}{5}+\dots +\frac{1}{2010-4k}-\frac{1}{2013-4k})=\frac{2}{3}(1+(\frac{1}{6}-\frac{1}{5}+ \dots + \frac{1}{4k-2}-\frac{1}{4k-3}+\frac{1}{6}-\frac{1}{5}+ \dots + \frac{1}{2010-4k}-\frac{1}{2009-4k})-\frac{1}{4k+1}-\frac{1}{2013-4k})<\frac{2}{3}<\log 9$](http://latex.artofproblemsolving.com/4/5/5/455c248c0b51905edb2480933600a251e1ae9a00.png)
,
and 
with such 
would repeat in both polynomials 
,
,
is real solution. This narrows down our search to all 
,
and 
for all positive integers 
,
and 
for all positive integers 
on the board.
In the case where the LHS is positive, the RHS can be positive or negative. If the RHS is negative, then the inequality is true. If the RHS is positive, notice that 
.
gives the desired.
,
if and only if 
.
for positive integer 
away from each other, so at most one of them can be in the interval ![$[-0.5, 0.5]$](http://latex.artofproblemsolving.com/c/1/3/c138148833360f097a3e1bbdf7c814a55881a9eb.png)
.
We define 
.![$[-1.5, 1.5]$](http://latex.artofproblemsolving.com/1/9/1/191d43b329600a0b628c2a21714aff30e6f52861.png)
has size 
,
This means that, in the expression we are trying to prove, 
the RHS is negative and the LHS is nonpositive. It would be sufficient to prove that 
This can be rewritten as 
We have previously obtained 
.
.
Then, this results in 
Our goal is implied by 
Furthermore, this is equivalent to 
Wow, I can't think of a "morally correct" way to prove this. Instead, let's do it this way:
and 
.
we can see that 
Thus, our goal is implied by 
.
allows us to see that 
,
