Math Puzzle [num 20] - Page 3

1)50 white and 49 black in one jar, 1 black marble alone in the other, equals almost 75 %.. like 74 something
part two is same answer

Put all 100 marbles into 1 jar.

But then you get to the part where you "pick a random marble from that jar" and if it's the empty jar your head explodes because you're a 1st generation android.

Let's try the following general strategy:

We start from a height called h(1). If the egg breaks we try all floors from 1 to h(1)-1. If it doesn't break we chose a different floor, h(2)>h(1). If it broke try everything from h(1)+1 to h(2)-1. The same way if the egg broke at h(i) we try every floor from h(i-1) to h(i)-1.

So if the first egg broke at h(k) our worst case scenario is the second egg breaking at h(k)-1, which would lead to k+[h(k)-h(k-1)]-1 steps. Since the problem asks us to find the best worst case scenario let's just chose the function h such that the expression k+h(k)-h(k-1) doesn't depend on k. That's just a series where the difference between subsequent elements decreases by 1. E.g 5, 9, 12, 14, 15

We know that the last floor number in the sequence h(l) is 100 and h(0)<0. Solving for n(n-1)/2)>100 n>14. So if the egg never breaks we need 14 tries, starting with floors 9, 22, 34 etc. The choice of the first floor could be anything between 9 and 14, it doesn't change the final tally.

Both buckets must have the same amount of both substances in the end, assuming a V/V basis and ignoring real-world effects.

This is because w1 + w2 = 1 = j1 + j2 ...(1)

Where w = water and j = JD.

We are told that in the end w1 + j1 = 1 = w2 + j2 ...(2)

His claim is that w1 > j2. However, we can show that any case where w1 != j2 cannot occur.

If w1 > j2, then from (1) we know that j1 > w2.
But if j1 > w2, then we get a contradiction at (2) because w1 + j1 > w2 + j2 even though they're supposed to be equal.

Another way to solve it is to plug (2) into (1) and get:

w1 + j1 = 1 = w1 + w2
j1 = w2

And likewise we can prove that j2 = w1.

I have actually not given you enough information to completely solve the puzzle. What bit of information did I intentionally leave out? What is the solution for all variations of this extra information?

Note: I may have unintentionally left some information out, but I've tried not to.

Note 2: It is possible to solve this puzzle without this extra piece of information (since I assumed not everyone would try the advanced version) - it's just that there will be multiple valid solutions. Hence the advanced version is essentially to identify what this missing information is and find all solutions. It is best to solve the simpler version first.

Did this one before... Start from the case of 2 pirates and work your way up:

• For 2 pirates, senior takes 100 gold, junior can't do shit.
  
• For 3 pirates, senior gives 1 gold to the lowest rank, and takes 99 gold for himself. Lowest agrees because if not, then senior is killed and we end up with the 2 pirate version proposal in which the lower rank gets no gold.
  
• For 4 pirates, senior gives 1 gold to 2nd lowest ranked pirate, and takes 99 gold for himself. Second lowest agrees because if not, senior is killed and we end up with the 3 pirate version proposal in which the second lowest gets no gold.
  
• For 5 pirates, senior needs 2 partners to corroborate, so he gives 1 gold to lowest and 3rd lowest rank, and takes 98 gold for himself. They will accept because if not, they will revert to scenario with 4 pirates in which they get no gold.
  
• For 6 pirates, senior needs 2 partners so he gives the 2nd lowest and 4th lowest 1 gold each.
  

This continues on and can be generalized for infinite number of pirates as:

• if there are an even number of pirates -> the senior pirate gives 1 gold to every pirate who is an even number'th rank from the bottom, and the rest to himself.
  
• if there are an odd number of pirates -> the senior pirate gives 1 gold to every pirate who is an odd number'th rank from the bottom, and the rest to himself.
  

I'm not sure what you left out; I'm guessing it's the number of actual gold coins to distribute, but I'm not sure why you would... this will bug me tonight...

anyway, assuming the thing u left out is indeed the number of coins, i'll call that quantity x, and the number of pirates y. General solution is:
Senior partner gets n-(y-2+y%2)/2 coins and the rest of the pirates who fit the rules stated above get 1 coin each.

This only trick I can think of is that the cube must be hollow.

Then to get the number of cubes, we take the surface area, subtract 1 for each edge, and add 1 for each corner.

n = 2 * (95*85 + 95*2000 + 85*2000) - 4 * (95 + 85 + 2000) + 4 = 727,434

The surface area is simply the first term in that equation, so we just take that over 6n (the number of faces) and get:

P = 16.9%.

Or roughly 1/6 since the 2nd and 3rd terms are tiny compared to the 1st.



EDIT: I just realised there are 8 corners, not 4. Oh well.

16150000 cubes in total

To get the number of cubes that could have been painted, we take the surface area of the cube before it got broken.
16150 = 95 * 85 * 2 (there are two faces with this size)
340000 = 2000 * 85 * 2
380000 = 2000 * 95 * 2
736150 = total

Each face has two instances of overlap, while the 8 corners of the cube have 3 instances of overlap.
727422=736150 - 4(95) - 4(85) - 4(2000) - 8, this should give the amount of cubes with a black face.
I think subtracting 8 there was probably an error.

Now we need to determine the amount of cubes that have 3, 2, and 1 possible painted side.
8 cubes have 3 sides painted by definition of a cube.

If my math up top is correct, you could subtract that number from the total painted cubes and arrive with the correct figure. I don't think it is, so we can:
8696=4(95-2) + 4(85-2) + 4(2000-2). This should give us the amount of non-corners.
To get the rest I'll use my probably flawed math from up top.
718718=727422 - 8 - 8696

# of sides : # of cubes
3: 8
2: 8696
1: 718718

Now onto probabilities
# of sides : Probability of rolling a black face
3: 50%
2: 33%
1: 17%

If I didn't already fail I'm really going to fail now.
Total probability = probability of picking a cube with black * probability of it being rolled and coming up black.
x= total cubes = 16150000
answer = .5(8/x) + .33(8696/x) + .17(718718/x)
answer = .0000002 + .00018 + .007 (didn't round up because it is .007 :D)
answer = 0.72%

If we're talking amnesia, the monster doesn't swim unless Day9 is in the water. The way to prolong his life is to go into the water on the opposite side of the island of the gate. Lure the monster there, then run straight across the island and to the gate. The water monster will have to run around half the circumference - a tangent line towards Day9 if he still hasn't reached the gate.

Even if the monster knows where Day9 is, the fastest path to him until he reaches the center of the island is still the radius line. This will keep the monster as far from the gate as possible until Day reaches the center of the island and moves off the center towards the gate. Only then will it start to circle around.