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+2 w
+1 w
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. .
can anyone confirm?
I did that like one hundred times...
and give up when evaluating the first sum.
