[/quote]
,
,
is chosen on the side 
.
and 
,
and 
are drawn, respectively. The point 
is the circumcenter of 
.
and 
lies on the line 
.
be a point in the plane of triangle 
such that the segments 
,
,
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to 
is acute.
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray 
and ray 
.
be a point on segment 
.
be the intersection of lines 
and 
be the intersection of line 
is tangent to 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of 
,
denote the incenter and 
.
perpendicular to 
intersects the tangent from 
.
and 
.
.
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than 
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
.
,
,
,
equals 
such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
numbers, none of which appears more than 
times. Prove that they can be divided into 
groups of 
be a cyclic quadrilateral. 
be the projections of 
,
is a kite.
.
and 
,
,
.
such moves the blackboard will contain a single number. What is its maximum possible value? (Reminder that 
is the maximum integer not exceeding 
.
Y by ManuelKahayon, Davi-8191, Wikson, mathleticguyyy, Adventure10, Mango247, deplasmanyollari, Marshall_Huang, NicoN9, wizixez, HWenslawski
Let 
be the number of positive divisors of 
. Let 
be the number of positive divisors of which have remainders 
when divided by 
. Find all positive integral values of the fraction 
.
be the number of positive divisors of 
. Let 
be the number of positive divisors of which have remainders 
when divided by 
. Find all positive integral values of the fraction 
.
for any 
not multiple of 
for any 
,
is a divisor of 
for some positive integers 
.
and 
for some prime 
and 
be the set of divisors of 
be the set of divisors of 
.
.
be the set of divisors of 
Suppose that 
or 
where all prime divisors of 
are type 1. Then 
and 
.
.
.
for some 
prime of type 2, 
and 
or 
and 
and 
.
and 
.
then 
and 
.
and 
.
then 
and 
and 
.
.
.
with 
or 
where 
.
since 
or 
,
.
and 
are coprime, then 
.
divides 
where 
,
coprime with 
are 
where 
,
,
and 
,
'
'![\[
\lambda:=\frac{\tau(10n)}{\tau_1(10n)}=\frac{(\alpha+1)\prod(\beta_i+1)\cdot \tau(m)}{\prod(\beta_i+1)\cdot \tau_1(m)}=(\alpha+1)\frac{\tau(m)}{\tau_1(m)}.
\]](http://latex.artofproblemsolving.com/f/3/8/f3880622c6a839318dfd2bc6910e46c27e2170b6.png)
Use generating functions. Define 
.
,
,
of 
is 
is even, and ![\[
G(1)=\tau(m),\qquad G(-1)=\tau_1(m)-(\tau(m)-\tau_1(m)).
\]](http://latex.artofproblemsolving.com/d/d/4/dd4d0ec976a5c72d91a4011f0d3d89725c6ddc22.png)
Now 
is 
if some 
is odd, and 
,
runs through the even positive integers. In the second case, 
is coprime to 
,
.
runs through the composite odd numbers due to 
)
be the prime factorisation of 
and 
be a divisor of 
then 
where 
and 
Therefore, the number of ways of choosing 
is the sum of the even coefficients of 
which is given by
![\[\frac{\tau(10n)}{\tau_1(10n)}=\dfrac{(b+1)(a+2)(c+2)\prod_{i=1}^{r_1}(\alpha_i+1)}{\left\lfloor\frac{\left(\prod_{i=1}^{r_1}(\alpha_i+1)\right)(a+2)(c+2)+1}{2}\right\rfloor}\]](http://latex.artofproblemsolving.com/2/e/0/2e0d34211ce1cbd4a0a15e0d6a9f7e41c27e7120.png)
If 
is odd then,![\[\frac{\tau(10n)}{\tau_1(10n)}=2(b+1)\]](http://latex.artofproblemsolving.com/b/3/f/b3f556bce406bd998cfa697caa5d0d257d436afa.png)
Hence, every even number is attainable by 
![\[\frac{\tau(10n)}{\tau_1(10n)}=\dfrac{2(b+1)(a+2)(c+2)\prod_{i=1}^{r_1}(\alpha_i+1)}{\left(\prod_{i=1}^{r_1}(\alpha_i+1)\right)(a+2)(c+2)+1}\]](http://latex.artofproblemsolving.com/7/b/8/7b8933224ee9601ed7e371c27a753da9d0e7541e.png)
Let 
and 
then we know that 
which means 
since 
Also 
is even, so 
for some 
Hence![\[\frac{\tau(10n)}{\tau_1(10n)}=k(a+2)(c+2)\lambda\ge 4\]](http://latex.artofproblemsolving.com/a/7/9/a79fdd681eb723713288187a9ea84cc6f150933a.png)
which is odd. Therefore choosing 
we obtain that every composite odd number 
is attainable by 
Where, 
and 
.
is not affected by 
.
are formed by either two divisors 
be a function that counts divisors 
We claim that 
This is easy to establish when 
.
.
be another distinct prime divisor 
.
(distinct than 
)
increases by 
.
forms the required divisors by interacting with the divisors 
And,
Using the above formula, it is easy obtain 
It's easy to observe, the above fraction can attain all even values when 
is even. Suppose, it is odd, ie, Let 
Now notice, 
must be odd and 
'
can attain any value. Also, 
and each of 
,
.
.
and 
equal 
and 
,
Let 
where for all 
which is easily proved by the trivial facts that if 
and 
Now given that 
let’s try to find 
Note that 
which, through cancellation by parity, gives 
Thus, using our base case, our induction is complete.
(raised to any power) is multiplied in 
if at least one of 
is odd. (which can be proved very similarly to our proof to Claim #1, by using induction with a base case and parity cancellations) Also, it is worth noting that 
since if a factor is multiplied by any power of 
which gives 
as long 
and 
both which give 
respectively.
where 
which still gives 
where 
is even for all 
which gives 
which is only an integer when 
is a multiple of 
by using the fact that 
and 
are trivially relatively prime by using our formulas obtained for them earlier, which generates multiples of 
)
and we are done.
.
,
.
doesn't change 
.
,
,
is a positive integer. Then 
Proof: Observe that a divisor 
(where we allow 
to be zero) of 
is even. First, suppose at least one of the 
.
.
arbitrarily. Then we have exactly 
choices of 
,
.
of 
.
in increasing order, starting with 
and ending with 
.
Now we compute the possible values of 
is not a perfect square and 
Using 
which covers all evens.
We can verify with the Euclidean Algorithm that 
,
is an integer, it must be divisible by 
.
where for 
,
and 
are even numbers that are at least 
,
where all 
and all 
and 
must be of the form:
Proof. Note that a divisor 
is 
iff it is the product of an even number of 
counting multiplicity. As a result, consider the generating function
Then the number of such 
is the sum of coefficients of all even power of 
in 
In particular,
Proving the claim. 
Where since 
and 
are all even, 
and setting 
So all odd composite integers are possible. If not all 
then
Whence all evens are possible, finishing the problem. 
where ![\[a=p_1^{a_1}p_2^{a_2}p_3^{a_3}\dots p_i^{a_i}\]](http://latex.artofproblemsolving.com/a/d/2/ad257832d334d34b26d3a170c379b4d945cc0ba8.png)
where 
are distinct primes equivalent to 
in the prime factorization of 
Similarly, define ![\[b=q_1^{b_1}q_2^{b_2}q_3^{b_3}\dots q_j^{b_j}\]](http://latex.artofproblemsolving.com/0/a/a/0aa06d7342c763eb74709def08810f207d2f71cc.png)
where 
are distinct primes equivalent to 
Now, ![\[\tau(x)=(p+1)(a_1+1)(a_2+1)\dots(a_i+1)(b_1+1)(b_2+1)\dots(b_j+1).\]](http://latex.artofproblemsolving.com/7/0/7/7072d24ad68ccf5469ca39bc9e21397e1e1b3ec2.png)
When picking a divisor that is equivalent to 
we cannot pick any factors of three. We must pick an even number of prime divisors of 
Thus, 
In particular if a number 
is a possible value, any multiple of 
if we have any 
Thus, no matter how we pick the other prime factors, there will always be half of the ways to pick the number of 
so 
in this case. Thus, all evens work.
where 
is an integer, so any multiple of 
is a possible value. Since 
so 
was a multiplicative function as that is the first thing you'd do in any textbook about arithmetic functions. Unfortunately I quickly found that it is in fact not multiplicative but has a property that I'll call 
.
where 
,
and 
is either 
.![\[ \textbf{Lemma 1: } \tau_1(XYZ) = \tau_1(X) \tau_1(Y) \tau_1(Z) \]](http://latex.artofproblemsolving.com/f/4/4/f4498abd2211f53162fae7373ccede4fc0fed7ea.png)
.
.
.
.
where both 
and 
and vice-versa so the number of possible 
is equivalent to the number of posible 
multiplied by the number of possible 
so our lemma is proved.
itself is multiplicative, we also have 
.![\[ \frac{\tau(n)}{\tau_1(n)} = \frac{\tau(X) \tau(Y) \tau(Z)}{ \tau_1(X) \tau_1(Y) \tau_1(Z)}\]](http://latex.artofproblemsolving.com/c/8/b/c8bf7cddaea604528936d7629da1acd7e94674f4.png)
.![\[ \frac{\tau(Y) \tau(Z)}{\tau_1(Y)} \]](http://latex.artofproblemsolving.com/8/3/6/836b719ec0bde34faeb9ec70bf5943b196695781.png)
,
and 
.
.
we would be able to generate 
hence our first observation is that all even numbers are solutions. Now consider the case of 
.
.
or 
.
so it would never generate odd primes. If we had the second case then to make the fraction an integer, all factors of 
would need to cancel with 
so we would end up with some integer multiple of 
and we already saw that this is never prime.![\[ \textbf{Lemma 2: } \tau_1(Y) = \lceil \frac{\tau(Y)}{2} \rceil \]](http://latex.artofproblemsolving.com/b/f/5/bf5a5dfb3083b446364e1a4fa875966ae408ce98.png)
.
and then defining:![\[ Y \to \left( \begin{matrix}
p_1^0 & p_2^0 & ...\\
p_1^1 & p_2^1 & ... \\
... & ... & ... \\
... & p_2^{a_2} & ... \\
... & ... & ... \\
p_1^{a_1} & 0 & ...
\end{matrix} \right) \]](http://latex.artofproblemsolving.com/0/b/f/0bf1a5d5757459d1efe8b6fe8d14d072cf63685b.png)
,![\[ \left( \begin{matrix}
1 & 1 & 1 \\
2 & 5 & 11 \\
4 & 25 & 0 \\
8 & 0 & 0
\end{matrix} \right) \]](http://latex.artofproblemsolving.com/5/2/d/52d1a335c9b58f6b48912a44af74a3de08179f7f.png)
as the amount of ways of picking one non-zero element from each column and then multiplyign them. So for the matrix above 
.
.
as the amount of ways to pick one non-zero element from each column, multiplying them, and then having the result be congruent to 
.
we can also just apply 
,
,
.
by induction on 
.
.
in which case 
and 
.
in 
would count for ![\[M = \left( \begin{matrix}
1 & & \\
-1 & M' & \\
1 & & \\
... & &
\end{matrix} \right) \]](http://latex.artofproblemsolving.com/e/c/c/eccd3de66021f124f2b4f470d93be90a9f73e222.png)
to the right. To calculate 
because we can take all of the products in 
in the column we would find the products in 
.
.
.
and this result is also consistent with our hypothesis.
but again 
.![\[\frac{|M'|(n-1)}{2} + \lceil \frac{|M'|}{2} \rceil = \lceil \frac{n|M'|}{2} \rceil \]](http://latex.artofproblemsolving.com/0/c/9/0c9d26cc0c040cf3b9b0fc6d53ff44cad84912cb.png)
but 
.
we can simplify this to 
which is a well known inequality. Hence the result is proved.
.
.
because of a number 
and hence 
.![\[ \tau_1(n) = \tau(X) \lceil \frac{\tau(Y)}{2} \rceil \]](http://latex.artofproblemsolving.com/6/d/d/6dd474037ac39fcf0ce54708cff388d1baf28570.png)
where 
and 
.
which is the sum of coefficients of 
which is
.
is even, then the fraction is 
which is even
.
.
and 
are greater than 
is composite.
.
,
,
,
且为素数.
为 
正因子当且仅当 
.
,
.
.
.
.
.
的情形, 则 
,
两个, 即 
.
,
,
,
均为偶数, 则 
.
,
.
可取遍全体奇合数.
where all 
and all 
for all a is positive integer or 
so we have proof for all composite numbers. If 
and it covers all even numbers.
So rest of the solution is easy for equality don't works for prime 
,
Plugging in 
where 
since 
divisors. Note that since 
are precisely the odd composite numbers (since 
If 
or 
,
(possible since 
.![\[r(n)=\frac{\tau_1(10n)}{\tau(10n)}.\]](http://latex.artofproblemsolving.com/d/3/4/d34c0ba50a91da4d74b09c7c83dae4e993826d80.png)
let 
be the set of possible values of 
.
we can get any multiple of 
appears. Otherwise use recursion to calculate the denominator; the number of prime factors cannot decrease and yet is at least two upon considering just two prime factors of 
and 
,
.
covers all the positive integral values.
.
,
and gcd
)
.
)
where 
.
.
be the divisors of 
.
are
and 
.
,
and
.
,
,
and
can attain all positive integral values except 
so 
for any positive integers 
Therefore, taking some composite number 
if 
then we get 
Hence 
where 
are positive integers with 
and 
s
Then 
while obviously 
If 
with 
dividing it with 
it is possible to correspond 
or 
if 
is odd or even respectively, making 
so 
is always even.
This yields 
so 
cannot possibly divide 
always divide 
is always composite. This makes 
is an integer it cannot be an odd prime.
