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or 
tile.
be an integer. Dominoes are placed on an 
board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some 
such that each row and each column has a value of 
. Prove that a balanced configuration exists for every , and find the minimum number of dominoes needed in such a configuration.
be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer 
such that 
. such that 
be the circumcircle of triangle 
with center 
, and the 
inmixtilinear circle is tangent to 
at 
respectively. 
is the intersection of 
and 
and 
is the intersection of 
and . Prove that the isogonal conjugate of lies on the line passing through the midpoint of 
and .
![$f \colon [0,1] \to \mathbb{R} $](http://latex.artofproblemsolving.com/2/1/8/21859d37ed035abd07d0be112f49d52598fa477e.png)
be a differentiable function such that its derivative is an integrable function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
, and 
. Prove that ![\[ \int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2 \]](http://latex.artofproblemsolving.com/2/3/e/23e33eed39fc1e20486bbb86261d3566e8fd2017.png)
be an odd prime number, and be an odd number not divisible by . Consider a field 
be a field with 
elements, and 
be the set of elements of 
, whose order is not in the multiplicative group 
. Prove that the polynomial 
has at least coefficients equal to 
.
and 
be a continuous function for which there exists an antiderivative 
, such that 
, for any 
, and
holds for any 
. Prove that 
for all .
be a positive integer, ![$g \in \mathbb{R}[X]$](http://latex.artofproblemsolving.com/0/1/e/01e7ebcebcb9f16935716ee859f7acfaa66c29d4.png)
, 
be a polynomial with all of its roots being real, and a polynomial function such that 
for any . Prove that 
for all .
be two matrices such that 
. Prove that:
is odd, then 
;
, then .
, the following 
statements are equivalent:
is differentiable, with continuous first derivative.
and for any two sequences 
, convergent to 
, such that 
for any positive integer , the sequence 
is convergent.
matrices which has only 
and for their entries?
, with their second derivative being continuous, such that the following holds for all 
: ![\[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]](http://latex.artofproblemsolving.com/0/2/9/029a935e28ca0a9b36e65702689cca91eeb8f614.png)
![$f\colon[0,1]\rightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/f/d/b/fdb1fb0216ab9d17007b1b7b03d20d1060b88531.png)
be a continuous function. Suppose that for each 
, the function ![\[f_t\colon[0,1-t]\rightarrow\mathbb{R}, f_t(x)=f(x+t)-f(x)\]](http://latex.artofproblemsolving.com/4/f/c/4fc6a66d0600a999c000136a26ea23dfcc269cc3.png)
has an unique global minimum point, which we will denote by 
. Prove that if 
, then 
is constant zero.
and be positive integers. The square in the 
th row and 
th column of an -by- grid contains the number 
. For which and is it possible to select squares from the grid, no two in the same row or column, such that the numbers contained in the selected squares are exactly 
?
is a nonconstant polynomial with integer coefficients with the following property:
and 
are both odd.
is a nonconstant polynomial with integer coefficients with the following property:
and are both odd.
where m has a large v_2 to get a contradiction.
. Note that no 
can equal zero since infinitely many primes divide zero. Also note that we cannot have 
for any positive integer . This is because its true for 
since they're odd and for 
, 
and 
which is a contradiction since both and 
cannot be odd.
is unbounded and also eventually 
is strictly increasing (for say 
)., assign to it the maximal prime power dividing and call it 
. Then clearly since only primes can divide it, this means ![$g(i) \geqslant \sqrt[k]{a_i}$](http://latex.artofproblemsolving.com/b/8/c/b8ca97c2ca3b1adc8b631cdead48414630f1558d.png)
.
be a sufficiently large number (bigger than 
) and consider the numbers 
. By PHP some two of these, say 
and 
are powers of the same prime. This means that 
must divide both 
and 
.
implies that 
for big enough and .
. But on the other hand we must have ![$\ell \geqslant \sqrt[k]{a_{N+i}}$](http://latex.artofproblemsolving.com/1/8/9/189e15c3462aaaea11c034aaf585caa3efb15582.png)
. So this means that 
for all sufficiently large , which fails for size reasons since is fixed.
and are both odd, this cleverly tell us all 
are odd and so it can't have an integer root.
, also let 
.
such that if 
then ![\[g(n)=i \text{ such that } \max(p_1^{e_1},\cdots,p_\ell^{e_\ell})=p_i^{e_i}\]](http://latex.artofproblemsolving.com/5/5/c/55cc2641e7c78ba1cd42ae607d9fa681a1a48724.png)
By PHP, there exists infinitely many 
such that 
. Say 
. So ![\[p^{\nu_p(a_n)} \ge (f(1) \cdots f(n)+1)^{\frac 1 \ell} \ge 0.99 c^n (n!)^{\frac d \ell}\]](http://latex.artofproblemsolving.com/d/6/3/d63595b0b18949cd00f382dfcd39482acbb62d6e.png)
WLOG assume 
(doesn't matter) and so ![\[p^{\nu_p(a_n)} \mid f(1) \cdots f(n)[f(n+1) \cdots f(m)-1] \iff p^{\nu_p(a_n)} \mid f(n+1) \cdots f(m)-1\]](http://latex.artofproblemsolving.com/f/b/4/fb45c8c2815926e079e3f22c1d581875f5b1012a.png)
And so we have ![\[p^{\nu_p(a_n)} \le f(n+1) \cdots f(m)-1 \le f(n+1) \cdots f(n+\ell) \le 1.01 c^\ell [(n+1) \cdots (n+\ell)]^d\]](http://latex.artofproblemsolving.com/8/b/2/8b26152ecf3230e3b4f84e6fe91b4a3ca5131e86.png)
Multiplying by 
both sides of the inequality and taking log we get 
Where we used Stirling's Approximation at the end.
but RHS 
and this is false for large enough .
and write 
.
are both odd, each 
is odd and hence as a result, each 
it holds that 
is a power of 
and 
.
.
(after all 
)![\[f(1)f(2) \dots f(2^v N) = \prod \limits_{0 \leq k < 2^{v-u}} f(kN+1)f(kN+2) \dots f(kN+N) \equiv \left[f(1)f(2)\dots f(2^uN) \right]^{2^{v-u}} \equiv (-1)^{2^{v-u}} \pmod p\]](http://latex.artofproblemsolving.com/a/5/7/a57896a6a1ce6ac86887e2d669702a94b777d664.png)
Since 
,
is even and this yields that 
which forcing 
,
for 
.
are powers of two. Replacing 
,
.
for some positive integer sequence 
.
if 
for each 
and hence we have 
.
.
where 
and 
is some constant. As a result, 
is 
.
we can simplify this to 
.
where we view them as functions of 
limits to 
for all 
.![\[f(M+1)f(M+2)\dots f(2M) - f(1)f(2) \dots f(M) > 10^{100}\]](http://latex.artofproblemsolving.com/5/e/8/5e8acddb36c1acd683e470355ee936368ea4bd50.png)
which implies that for large enough 
.
,
where 
suffice large such that 
,because there is non of 
is divisible by 
because 
where 
adsurb.