AoPS Academy
points are marked, no three of which are collinear. All possible segments between them are drawn. Grisha assigned to each drawn segment a real number with absolute value no greater than 
. For every group of 
marked points, he calculated the sum of the numbers on all 
connecting segments. It turned out that the absolute value of each such sum is at least 
, and there are both positive and negative such sums. What is the maximum possible value of ?
with domain ![$[0,2025]$](http://latex.artofproblemsolving.com/d/1/7/d17ac41d54c4b1888fb8bb43cf11b372691694a0.png)
satisfying 
and 
is strictly increasing on Compute smallest real M such that all functions in this set ,
.
have local extrema at 
respectively. find 
. Here 
are constants .
be the region in the complex plane enclosed by curve 
for 
. Compute perimeter of .
(0<x<2π)
if 
is even.
a positive integer . suppose that there exist two automorphisms 
of 
such that 
. is even.
be a positive semi-definite matrix. Prove that the matrix 
, is also positive semi-definite for all 
.
?
such that 
for all positive integers .
and 
be two points that are diametrically opposite to each other on a unit sphere. right square pyramids are fitted along the line segment 
, such that the apex and altitude of each pyramid 
, where 
, are 
and 
respectively, and the points 
are collinear. pyramids, with altitudes of equal length, that can be fitted in the sphere, in terms of . pyramids that can be fitted in the sphere, in terms of . tends to infinity.
The last line only since I like these variables better, heh. Let 
[which is called the Liouville Function] then we can see that 
is multiplicative, as 
. But by Dirichlet Convolution ![\[ \sum_{d \mid n} \lambda(d) = \lambda \ast \textbf{1} \]](http://latex.artofproblemsolving.com/8/3/f/83f7786937968849d11e513011352975cbddf8ea.png)
we can see that 
is also multiplicative. So it suffices to look at prime powers 
. It's now easy to conclude that ![\[ \sum_{d \mid n} \lambda(d)= \begin{cases} 1 \text{, if } n \text{ is a perfect square}
\\ 0 \text{, else.} \end{cases}\]](http://latex.artofproblemsolving.com/6/9/a/69ac3eb71aac8e945da61a5fc46e0917a77ea671.png)
So it suffices to count the number of perfect squares smaller than 
and well 
, hence the answer is 
.
is multiplicative. We have that![\[\displaystyle\sum_{n=1}^{1989}{(-1)^{\Omega(n)}\lfloor\frac{1989}{n}\rfloor} = \displaystyle\sum_{n=1}^{1989}{\displaystyle\sum_{n\mid m, m\leq 1989}{(-1)^{\Omega(n)}}}\]](http://latex.artofproblemsolving.com/1/d/0/1d08d012ecc2a91a2274fd6d81ff99dda96e42e3.png)
We can switch the order of the summation to make it![\[\displaystyle\sum_{m=1}^{1989}{\displaystyle\sum_{n\mid m}{(-1)^{\Omega(n)}}} = \displaystyle\sum_{m=1}^{1989}{\textbf{1}*(-1)^{\Omega(n)}}\]](http://latex.artofproblemsolving.com/6/0/3/603147c02250d726a594c2f7acb538c0d9704ef6.png)
Because is multiplicative, 
must also be multiplicative. For a prime power 
, note 
is if 
is even, and 
if is odd. Thus, we have that is if is a square and otherwise.![\[\displaystyle\sum_{m=1}^{1989}{\textbf{1}*(-1)^{\Omega(n)}} = \lfloor\sqrt{1989}\rfloor = \boxed{44}\]](http://latex.artofproblemsolving.com/4/b/7/4b7ce3a04966a30cd2da183013ebe57c19dd4110.png)
Now let 
where 
are distinct primes and 
are positive integers. Since is mutiplicative, we have
If at least one of the is odd, the sum is . On the other hand, if all are even, the sum is . Then the final sum counts the number of perfect squares between and inclusive, which is .
.![\[
\sum_{n=1}^{1989} (-1)^{\Omega(n)}\left\lfloor \frac{1989}{n} \right\rfloor = \sum_{n=1}^{1989} \sum_{m \mid n} (-1)^{\Omega(m)}.
\]](http://latex.artofproblemsolving.com/6/f/1/6f1df800004014b1683a01bb1b237f3268575286.png)
We will now show that![\[
\sum_{m \mid n} (-1)^{\Omega(m)} = 1
\]](http://latex.artofproblemsolving.com/9/2/2/9224382e59164762af692df149ec6d0c16cd8151.png)
if 
is a square and 
if is not a square. Notice that if is not a square, it has an even number of divisors, so we can partition the divisors of into groups 
. Notice that each group has sum because one of them has one more prime divisor than the other. If is a square, then we can do the same thing but with 
for some 
. Then 
, so the sum is 
. So summing from 
to 
is just the number of squares less than , which is .
Now see that the inner function is basically 
and both are multiplicative functions and hence their Dirichlet convulation is also a multiplicative function; and now we can check easily that ![\[\sum_{n \mid m} (-1)^{\Omega(n)}= \begin{cases} 1 \text{ if $m$ perfect square} \\ 0 \text{ otherise} \end{cases}\]](http://latex.artofproblemsolving.com/a/1/9/a19cebd6063504d98b3e0c8c93084b113117d03f.png)
This basically gives us that the answer is 
.
such that:
such that: 
.
such that
denote the number of prime factors of ![\[ \sum_{n=1}^{1989} (-1)^{\Omega(n)}\left\lfloor \frac{1989}{n} \right \rfloor. \]](http://latex.artofproblemsolving.com/2/0/4/2043dca9f2ff2adfd3c8d49fa4dd278871968db0.png)
It is often useful to rewrite the floor function in order to swap the order of summation, so let us do exactly that here:
Now, we will prove the following lemma that we obtained from checking small cases:![\[
\sum_{m \mid n} (-1)^{\Omega(m)} = 1 \iff n\text{ is a perfect square;}
\]](http://latex.artofproblemsolving.com/e/b/0/eb0f4b6a7767af51d40578ac00534ae8032681de.png)
![\[
\sum_{m \mid n} (-1)^{\Omega(m)} = 0 \text{ otherwise.}
\]](http://latex.artofproblemsolving.com/4/9/1/4915b23e26fb5c5e1ccf55e5614bca5eb64d8579.png)
Proof: 
- then we can easily see that it is multiplicative. Therefore, it suffices to check the function for prime powers 
if 
is even and 
if it is odd. Next, observe that since![\[ \sum_{d \mid n} f(d) = f \ast \textbf{1},\]](http://latex.artofproblemsolving.com/e/7/6/e7654598a9425ac392bee97751577393a5b1d2e0.png)
the function 
is also multiplicative. However, if a prime number 
has its 
odd, then we would have a zero - otherwise, we have a 
e
so the lemma is proven.
From here, since 
,
.