Math Puzzle [num 20] - Page 4

I think it's a "trick" question because you can overthink by looking at edges, corners and whatnot. There are 6x85x95x2000 sides and 2x(85x95+85x2000+95x2000) sides are painted black. There's no need to treat edges and corners differently: you want to double and triple count them because they have 2 and 3 sides painted black respectively.

Day9 would start to run in a spiral until he reaches the inner circle with radius length r/4. He would time make the spiral pattern in a way so that when he reaches r/4 away from the center, the monster would be on the other side of the island, lagging a full 180degrees behind. That is when Day9 would make a run for it straight to the coast (no more spiraling), and hopefully find a gate in the water. He can definitely make it to the coast, but not much more.

I think this is the best strategy because the max distance you can get the water monster to swim around is half of the island's circumference, or pi*r. At the circle with radius length r/4, you can keep up with the monster in terms of angular velocity (2*pi*r / 4s = (2*pi*r / 4) / s), so you can definitely outrun the monster inside that circle. Since Day9 is so smart, I will leave it up to him to come up with the spiral formula that will land him at the r/4 point with exactly a half circle lead on the monster. The distance Day9 covers from that point to the point on the coast along the same radial line is 3r/4, which will take 3r/4s time, while the monster must swim pi*r, which will take pi*r/4s, about .14r/4s more than it takes Day9 to reach the coast. If Day9 doesn't see a gate nearby, he must repeat the same process for every degree, minute and second of the island's entire 360 degrees.

Not sure if this is the solution you were looking for... something feels amiss

So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.

Let's consider the 50th cat.

It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.

We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.

Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.

What about cat 2? There is a new "bump wave" every 1 s, ...

Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.

Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.

So the last pair of cats fall off at 96 seconds.

So my generalization still holds for pirates <= 202, because the seniors would still want to live even if it means getting no gold. Now, I'm not sure how I would go about it afterwards... When n=203, senior needs 101 votes to not be killed, but only has 100 coins... T_T

I think the first egg need to go up the floor in increments of 3. This is the best solution I can think of.

The first thing I notice is that we need decreasing intervals between drops.

So if we start on 10, then we want both break and non-break scenarios to have similar remaining tries.

Break means we try 1->9, worst case scenario of 9 subsequent tries (10 total).

So 10 is only an optimum start if non-break also has a worst case of 9 subsequent tries.

And we know this is no longer possible because this means we then pick 19, then 27, and it's obvious we're not going to get very far.

Looking at the high end, we know that our last try is 100 to check if the eggs ever break. And we want to pick n such that we minimise the number of iterations before reaching 100.

We can solve this by working backwards.

We must end on 100.

2nd last is 99. It can't be 98 because if 100 breaks then we have possibilities [99,100] and no tries remaining.

3rd last is 97. That way, if 99 breaks, we can try 98 with our final try.

4th last is 94. This is because if 97 breaks, we have 2 tries left, for 95 and 96.

From here it's clear that the size of the spread increases by 1 for each try we go back.

This gives the sequence:

9, 22, 34, 45, 55. 64, 72, 79, 85, 90, 94, 97, 99, 100.

This involves a total of 14 tries in the worst case.

So the minimum number of tries is 14. We can begin on any floor [9,14]. My solution covers the lowest floors possible for each attempt in the primary sequence. The highest floors possible can be found by starting on 14 and increasing in increments of 13, 12, 11, etc. and ending on floor 100.

So you've worked out that he can safely make it to the water. But he's still got more tricks up his sleeves. What else can he do to increase the time he stays alive after reaching the water?

If you don't assume he runs on the line connecting the gate and the center of the island, and instead runs on a curve from the r/4 circle, he might get closer to the gate before the monster reaches him. Of course, exactly how much of a lead this would give him would diminish based on how far the gate is, but would still be >0 as long as the gate is not infinitely far away. I can't think of how to model this though, I think it would involve some sort of curve.

Don't consider collisions as bouncing cats. Instead, every collision has the same effect as having the two cats pass through one another. Thus, the time will be the maximum of the times it would take each individual cat to fall off if no other cats were present. We have a 99m pole and a cat on the left facing right moving at 1m/s, so this is 99s.