AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
+1 w
children sit in a circle facing inward with 
, and each child starts with an arbitrary even number of pencils. Each minute, each child simultaneously passes exactly half of all of their pencils to the child to their right. Then, all children that have an odd number of pencils receive one more pencil.
checkerboard satisfy the following conditions:
checkers have been placed on the board.
, 
is the Gergon point and the incircle 
touch 
, 
, 
at 
, 
, 
respectively.
, 
, 
be 
, 
, 
respectively. in 
and then we get the circle 
in 
and then the circle 
in 
and then the circle 
, , are coaxial.
and primes 
such that 
(there is at least 1 zero) can't be a perfect square.
such that![\[f((x\oplus f(y))+y)=(f(x)\oplus y)+y\]](http://latex.artofproblemsolving.com/d/3/5/d355f091797570bdc6c023eedb30e8ee9deac168.png)
Note: 
denotes the bitwise XOR operation. For example, 
.
is divisible by 2024.
be a scalene triangle such that 
is its incircle. is tangent to at . A point (
) is located on .
, 
, and 
be the incircles of the triangles 
, 
, and 
, respectively. and is also tangent to .
, let be the result.
, where 
is the Euler's totient theorem, which calculates the number of integers less than relatively prime to . again because why not, let the result be 
.
for 
. Let 
be the remainder 
leaves when you divide it by 
.
be your predicted USA(J)MO score.![$\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$](http://latex.artofproblemsolving.com/c/d/3/cd3f7d917b0ec944ec3017e1f737c538e1edd10e.png)
.
, and 
, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance 
away. What is the least number of moves that Carina can make in order for triangle to have area 2021?
in the coordinate plane where and 
are both integers, not necessarily positive.)
for which the sum ![\[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\]](http://latex.artofproblemsolving.com/6/b/0/6b010978a55642b7f7aabb2df2f7cf45c9db937a.png)
is an integer strictly between 
and 
. For that unique , find 
.
denotes the greatest integer that is less than or equal to .)
because each floor can only change the sum by less than . The natural choice is a value of near the one that makes the aforementioned sum exactly . Luckily, we compute![\[ \sum_{n=1}^{2023}\frac{n^2-na}5 = \frac15 \left( \frac{2023 \cdot 2024 \cdot 4047}{6} - a \frac{2023 \cdot 2024}{2} \right), \]](http://latex.artofproblemsolving.com/6/f/3/6f3a4b9c68dc61c013eb95a7935185bc5213d8dc.png)
which is zero at 
, an integer. Trying 
for the value of 
, we have![\[ \begin{aligned}
U=\sum_{n=1}^{2023}\left\lfloor\frac{n^2-1349n}5\right\rfloor &= \sum_{n=1}^{2023}\left(\frac{n^2-1349n}5 - \frac{(n^2 - 1349n) \mod 5}5 \right) \\ &= -\frac15 \sum_{n=1}^{2023} (n^2 - 1349n) \mod 5
\end{aligned}\]](http://latex.artofproblemsolving.com/a/e/4/ae49b2b8315784deabe534b96dda2e57f2ddd6a6.png)
But note that 
takes on the following values at the residues mod 
,
to 
passes through almost 
periods of length , only missing two zeroes that don't matter, with each period summing to , so![\[ U = -\frac15 (405 \cdot 5) = -405.\]](http://latex.artofproblemsolving.com/1/2/8/1285718100ceabc8bb7c5c93928f4c199b4e29c0.png)
It follows that 
.
. This means that![\[\displaystyle\sum^{2023}_{n=1}{\frac{n^2-an}{5}}\leq \displaystyle\sum^{2023}_{n=1}{\biggl\lfloor\frac{n^2-an}{5}\biggr\rfloor}< \displaystyle\sum^{2023}_{n=1}{\frac{n^2-an}{5}}+2023\]](http://latex.artofproblemsolving.com/1/1/d/11d9d7328fcff9ec31a2e1135a6f615c471dca1d.png)
Note that this means that 
must be otherwise any other value of would overshoot and . We use the sum of squares formula to get . Now, we engineer's induct on for the value of 
to get that it must always be 
. Thus, 
. Thus, our answer is 
.
for the value of to get that it must always be .
for the value of to get that it must always be .
. It follows that 
Using summation properties, we obtain 
Here, we observe that 
quickly grows/becomes extremely negative (as must be a positive integer), while the term 
does not relatively affect the size of on the same scale. Thus, 
. This means that 
so we must compute the sum of the fractional parts of 
for from to 
.
.
or 
then 
;
or 
then 
;
then 
;
then 
;
then 
;
then 
. to there are numbers 
, numbers 
, and numbers 
, so 
. Our final answer is 
.
.
, setting equal to (i.e. minimizing absolute value of the integral) and solving for yields 
, and rounding gives as the most likely answer. using your favorite method. 
.
. Approximate it as,
Then from our bounds on we find that 
should be clsoe to satisfying,
However then we approximately need,
Due to how enormous the middle term is we need 
. Now we can use this in our original floor equation to find,
Now the first sum is easy to evaluate using standard tactics. For the second term it suffices to do casework on modulo . If 
, we have 
. Similarly if 
we have 
, if 
we have , if 
we have 
and if 
we have . Then we have,
and our final answer is 
. We have
Now we can also deduce the weak bounds
The problem statement is simply saying 
. Here it is obvious that any value of where 
would place outside of the bounds, for any value of in the previous bound. So 
Then 
, and the fractional part cycles every , so 
yield 
respectively. Then from 
to 
the sum is 
, and 
yields an additional so 
. We obtain .
., setting equal to (i.e. minimizing absolute value of the integral) and solving for yields , and rounding gives as the most likely answer. using your favorite method. .
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and![\[ f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/c/8/9/c895ae7fdf8d7f284ac9fc94cc077d6edad6cbf0.png)
unless of your age can be written as 
.
can anyone confirm?
I did that like one hundred times...
and give up when evaluating the first sum.
