Calvinball!

Update: Now that everyone has had time to work on this problem by themselves, let's open up the floodgates and post your solution as a reply (or edit) to your comment. Let's see how many different ways the creative brains at Brilliant can approach this problem.

I used the Pick's Theorem, which states that in a Cartesian plane: $P(A) = i + \frac{b}{2} -1$, where $P(A)$ represents the area of the polygon, $i$ represents the number of interior points in the polygon and $b$ represents the number of boundary points.

First, notice that the $1$ pointer is determined by the number of $5$ and $13$ pointers. Then, I let there be $x$ 13-pointers and $y$ 5-pointers Then, draw lines $x=0$ and $y=0$ to form a triangle. The maximum number of $13$ pointers is $240$, while the max. number of $5$ pointers is $624$, hence, the vertices of the triangle are $(240,0)$ and $(0,624)$. [ The triangle is formed by drawing a best-fit line connecting$(240,0)$ and $(0,624)$]. See here that $b+i$ equal to the number of $(x,y)$ that satisfies our question. [plus some exceptions as we see later]

Here, we can easily derive the number of points lying on our best-fit line, which has a gradient of $\frac{13}{5}$. Notice that every $x$-coordinate which is a multiple of $5$ lies on the best-fit line. Hence, there are $\frac{240}{5}=48$ points lying on the best-fit line [not counting the point $(0,624)$. Then, $b = 48 + 624 + 240 = 912$.

Next, the area of the triangle is simply $\frac{1}{2} \cdot base \cdot height = \frac{1}{2}\cdot 624\cdot 240 = 74880$

Fitting the values of $b=912$ and $P(A) = 74880$ into Pick's theorem, we will get: $74880 = i + \frac{912}{2} -1$ Here, it is clear that $i = 74425$. So, $b+i = 75337$. But we're not done yet!

As mentioned earlier, we still have to add some exceptions! Notice that NOT ALL of the points $(x,y)$ is under the best-fit line! Some of them are above the line but still fulfills the conditions, and the Picks' Theorem fails to consider these points because they are not within the triangle.

Notice that the points with $x$-coordinates $2$ more than the $x$-coordinates of the point on the best-fit line has gradient more than $\frac{13}{5}$, which means they lie outside the triangle. For example, since $(0,624)$ lies on the lie, $(2,619)$ is above the line. (you can check it out) Since the cycle repeats itself in multiples of $5$, this is true for all points lying on the best-fit line. Since there are $48$ points (not including $(240,0)$ as it is the maximum number of $13$ pointers) on the best-fit line, there will be $48$ points that lie above the best-fit line that satisfies the question.

Hence, answer is $75337 + 48 = \boxed{75385}$.

P.S. Sigh... This is very tedious I think... Anyone has ideas to shorten this method?

Here's my solution:

The number of ways to score $3121$ is essentially the number of non-negative integral solutions to the Diophantine equation $a + 5b + 13c= 3121$. We note that any valid choice of $(b, c)$ determines a unique non-negative $a$, so we can simply afford to find the number of solutions to $5b+13c \leq 3121$. Rearranging this gives $b \leq \frac{3121-13c}{5}$ The number of solutions of $b$ for a fixed $c$, thus, is $\left \lfloor \dfrac{3121-13c}{5} \right \rfloor + 1$. We need to sum this up from $c=0$ to $c= \left \lfloor \dfrac{3121}{13} \right \rfloor = 240$, which gives our desired answer: $\sum \limits_{c=0}^{240} \left ( \left \lfloor \frac{3121-13c}{5} \right \rfloor + 1 \right ) = \boxed{75385}$

Does order matter?

About 40 mins, Without revealing too much, is there a faster way to perform discrete integration to find the number of points within an section of the cartesian plane?

I have the feeling that I shouldn't have listed them all out... 2 hours, though. Hehe...

Not sure if I did correctly... applied the method I have learnt from Score!... Took around 6.5mins...

It's simple... Its not very lengthy and tedious.So can we all share our solutions?

Well, i guess it could have been made much better if the possible points were $2, 5$ and $13$, now, since most of people have seen this post , are we allowed to post our solutions.

_Edit - _ Here is how i did:

Say we have $a$ number of 13- pointers, $b$ number of 5- pointers and $c$ number of q-pointers.

For a given $a$,

$5b + c = 3121 - 13a$,

Every value of $b$ gives one way, and $b$ varies from $0$ to $\lfloor \frac{3121 - 13a}{5} \rfloor$

Also , no. of ways for a given $a$ is $\bigg\lfloor \frac{3121 - 13a}{5} \bigg\rfloor + 1$, and $a$ varies from $0$ to $240$

Hence, total required number of ways = $\displaystyle \sum_{a=0}^{240} \Big(\bigg \lfloor \frac{3121 - 13a}{5} \bigg \rfloor + 1\Big)$

= $\fbox{75385}$

However, in calvinball, the rules can be changed .....on the spot...

Can I use a computer to calculate it? I found out how to find the answer, just can't calculate it by hand. :\

EDIT: oh yea, almost forgot. Took 4 minutes to find how to get the answer. Writing a computer program to find the actual numerical answer right now.

Around 4 minutes.I'm not sure I got the right answer though.

About 6 minutes, but I had to use a calculator at the end.

Around 11 minutes.

It took not more than 9-10 minutes. I'd developed the logic only in few seconds but manually computing the final value (which was a tedious task) took the whole time.

About 7-8 min.

it took me 10 minutes, but it was simple

Took 13 minutes.

4 minutes.but my ans is not within the range [0,999] :/