AoPS Academy
are group. If 
, prove that 
or 
. 
has at most one solution continuous for 
are complex numbers such that
and if 
denotes the sum of all products of these 
numbers taken 
at a time, prove that
whenever the denominator of the left-hand side is different from 
.
![\[\lfloor x \rfloor\]](http://latex.artofproblemsolving.com/d/3/f/d3f8facf62413e1f17c89b9c183a8735c2defe8b.png)
denote the greatest integer less than or equal to x . What is the sum of the squares of the real numbers x for which ![\[ x^2 - 20\lfloor x \rfloor + 19 = 0 \]](http://latex.artofproblemsolving.com/f/a/9/fa96f2d6217a16029e5f03499e0a83deffdfa28a.png)
of the pizza, where is a positive integer. Find .
Is there an elegant way to prove this?
is called stylish if the set 
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
. For example, 
is stylish because the set 
can be partitioned as 
, with sums , 
, and respectively. Determine the number of stylish numbers less than 
.
for any positive reals satifying 
?
to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
, but am confused on how to prove this statement for general vectors 
and 
although its supposed to be a quick comparison.
chance of landing heads when flipped. Krithik is playing a game where he starts with 
point. Every turn, he flips the coin, and if it lands heads, he gains point, and if it lands tails, he loses point. However, after the turn, if he has a negative number of points, his point counter resets to . Krithik wins when he earns 
points. Find the expected number of turns until Krithik wins.
has a pole at infinity then is a polynomial, but I can't conclude that 
is rational, how to proceed?
is a polynomial. Notice that, is a real-valued function and not holomorphic.
implies that has a pole at 
into itself. Using the compactness of 
you should verify that the number of zeros and poles are finite. Divide by product of all terms like, 
where 
is a zero or pole of order 
to kill off all the zeros and poles of and use the compactness and maximal modulus principal to show that the resulting holomorphic function is constant.
such that 
for all 
. (in fact this can be replaced with goes to infinity, a weaker hypothesis). This means that the zeroes are all inside the closed disk of radius 
. Since the domain of a meromorphic function is open and path connected(removing countably many points from the plane leaves a path connected set) the set of zeroes are discrete and closed hence finite. 
has removable singularities at the poles of , so it only has finitely many non-removable singularities (the zeroes of ) and in fact it is not hard to show they are poles. Hence we can write 
where 
is entire and 
is a polynomial. So we have that 
and so clearly 
for . Now we can show that is indeed a polynomial by estimating 
for 
where 
, estimate it to show that 
(hint: express as an integral and perform an argument similair to the proof of louville's theorem)
and 
Prove that
Prove that
![\[\mathcal{P}(x) = x^{108}+x^{102}+x^{96}+2x^{54}+3x^{36}+4x^{24}+5x^{18}+6\]](http://latex.artofproblemsolving.com/e/b/2/eb28918359848dd07784a3c2c6a34b934743c016.png)
be 
.![\[\dfrac{r_1^6+r_2^6+\dots+r_{108}^6}{r_1^6r_2^6+r_1^6r_3^6+\dots+r_{107}^6r_{108}^6}\]](http://latex.artofproblemsolving.com/3/1/5/315341a7adf35bc04e40ee0685a011c9f7b76115.png)
without Newton Sums.
so that ![$|f(z)|\xrightarrow[|z|\to\infty]{}\infty.$](http://latex.artofproblemsolving.com/e/5/8/e58ca320e7e5c0f372da23f1d836955a30573087.png)
Prove that 