Brainteasers/Math problems - Page 12

Ref:
[BS]: BullShit or Blizzard website, you choose which one I took this fact from

Isn't this one all about division to circuits? The prisoners must decide to enforce an order on the boxes. Therefor, they decide to randomly send each person to a different box (dubbed "his" box), and the guy opens the box, and goes to the box belonging to the name inside it, according to the what they defined as that person's box. Basically, you enforce an order of circuits here, the chances of a circuit being 51 or greater are very close to ln2 (combinatorics for defining the permutation, some calculus to calculate the limit of the series needed to explain this number). Therefor, the chance of success would be (1 - ln2)*100 percent, somewhere around 30%. Not quite sure though.

Incorrect phrasing. The manager has 25 dollars in his pocket, his son has 2, and each of the guys has one (out of the 30). No dollar unaccounted for. The trick is that they payed 27, 25 are with the manager and 2 are with his son. No dollar vanished.

Each domino has to be on one black square and one white square. You eliminated 2 of the same color, now there are 32 of one color and 30 of the other, therefor, it will be impossible to put the last domino, as the only two squares left are of the same color, therefor not next to eachother.

He pulls one out of the bag, and does something with it so that no one can retrieve it. Maybe he swallows it, and now it's in his digestive system. Maybe he throws it off the cliff before anyone gets a good look at it. The point is, everyone knows he just picked a ball, but they didn't see the color, and now they have no way of directly seeing it's color because the ball is gone. Lost. Irretrievable FOREVER!!!!!

So the only way they can determine the color of the ball is through logic. They open the bag and look at the remaining ball. It's black? Oh, then the one that he picked first MUST have been white!!!

Instead of thinking about the merging and reincarnating separately, think of the two combined as one process. So what are our options?
1.) Two high templar merge, and when that archon dies, it will reincarnate as a high templar.
2.) Two dark templar merge, and when that archon dies, it will reincarnate as a high templar.
3.) One of each templar merge, and when that archon dies, it will reincarnate as a dark templar.

If we perform "Action 1" on our initial group of templar, it is equivalent to removing a high templar from the group. Our new set is the initial set, minus one high templar. Cool.

If we perform "Action 2" on our initial group of templar, then our new set is the initial one, plus one high templar, minus two dark templar. Also cool.

If we perform "Action 3" on our initial group of templar, it has the same effect as Action 1. We just lose a high templar. Our new set is the initial set, again minus one high templar. SUPER COOL.

Notice the effect of each action in terms of parity. All three of these do not change the parity of the number of dark templar, and all of them DO change the parity of the number of high templar. So...

If we have an even number of dark templar to begin with, we will have an even number of dark templar at the end. Namely, we will have zero dark templar at the end, and our last templar is a high templar.

If we have an odd number of dark templar to begin with, we will have an odd number of dark templar at the end. Namely, we will have one dark templar. He will be the last templar, just sitting their chilling. Perhaps we could write a screenplay about him, much like "The Last Samurai," but in this case he'd be "The Last Templar." Of course, he would still be played by Tom Cruise.

Label and divide the coins in to three groups: ABCD EFGH IJKL.

Start off by balancing ABCD and EFGH
Three possibilities, and I will address the simplest ones first:
1) ABCD = EFGH, find the fake coin in IJKL in two tries, you should know how to solve this, weigh ABC against IJK, and then I against J if it doesn't balance.
2) ABCD < EFGH, this is the same as the next one, just switch labels

3) ABCD > EFGH
Weigh ABCEF against DIJKL, remember IJKL are now real coins and we have three possiblities:

1. ABCEF=DIJKL, we have now cleared A B C E F D by balancing the scale, we know one between G and H is fake and it weighs less than a real coin, balance G against H and find the lighter one.

2. ABCEF>DIJKL, D is clear because it was in the heavier group in step two, E and F are clear because they could only be lighter if they are fake. One of the A B C is fake and the fake coin is heavier, balance A against B and find the fake coin.

3. ABCEF<DIJKL, A B C are clear since they could only be heavier if they were fake. D E F are suspects here, either D is too heavy or E / F is too light. Balance E against F and find the fake coin.

No, yours doesn't work and the one you quoted works fine. The middle sister's answer is essentially random so you cannot ask a question that results in any possibility of marrying the person you asked. As a result, you have to ask a yes/no answer where yes = marrying one of the other two and no = marrying the other; if you get the middle sister the answer conveniently doesn't matter as both are eligible and thus the randomness is a nonfactor.

So, all you have to do is find a question which produces one answer for [NOT middle] and one answer for [middle] regardless of whether it's asked to the eldest or the youngest. The age question is such a question, because

-From the perspective of the eldest sister, she is the eldest and the liar is the youngest (which is the truth), therefore the middle sister is older than the liar.

-From the perspective of the youngest sister, she is the eldest and the truthteller is the youngest (the opposite of reality, since the youngest sister always lies), therefore she will also say that the middle sister is older than the liar. Remember that the youngest is compelled to lie and cannot choose not to, so her answer will be consistent.

So you ask sister A "is sister B older than sister C?".

-If sister A is the middle sister, either B or C are fine so the answer doesn't matter.
-If sister A is the eldest and B is the middle sister, she will answer "yes" so you should pick C (the younger = liar)
-If sister A is the eldest and B is the youngest, she will answer "no" so you should pick B (the younger = liar)
-If sister A is the youngest and B is the middle, she will answer "yes" so you should pick C (the older = truthteller)
-If sister A is the youngest and B is the eldest, she will answer "no" so you should pick B (the older = truthteller)

So we end up with

-If sister A answers "yes", pick C. Otherwise, pick B. This works for all permutations of A,B,C.

n lifeboats * n men per lifeboat = n^2 men

n-10 lifeboats after 10 sink * n+10 men per lifeboat = (n-10)(n+10) remaining men

drowned men = n^2 - (n-10)(n+10)

= n^2 - (n^2 - 10n + 10n - 100)

= 100

Uhm, isn't that why there is a binary expression for every number?

He could pick a black one, but then get them to check what was left in the bag. Since a black ball is left they would "know" he drew a white one and he would go free. And if things didn't go to plan he would at least expose that he was being cheated and might get a "fair" shot at avoiding justice. Not sure what any of this would have to do with where he is. I guess he could "accidentally" drop his black ball off the cliff to get them to check the bag.

If we say that the dark templar is 1 and the high templar is 0, we have a xor operation (see http://en.wikipedia.org/wiki/Exclusive_or )As we know, the XOR operation is commutative (the order doesn't matter) so we can merge in the order we want.
Let's say we have M dark templar and N high templar :
Let's merge all the dark templars together : if M is even, we get one high templar, otherwise we get one dark templar.
Let's merge all the high templars together : we get an high templar. (ie 0 XOR 0 XOR 0 XOR 0... =0)

Result :
If M is even, we get an high templar.
If M is odd, we get an dark templar.

Answer would be no. each domino piece would, by necessity, cover 1 black square and 1 white square. Because there are 32 white squares and 30 black squares, once you've placed 30 dominos, the 31st would have to cover 2 white squares, which is not possible.

The exact number of finite things there is to know?

Didn't actually think of that myself. There is another one.

The knowledge that he knows all there is to know?

then the last thing he'd know is that he couldn't know everything.

- sell 3-liter jug
- empty one 8-liter jug into pool 1
- fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them
- fill pools 3+4 with 4 liters of water each
- fill one 8-liter jug with the 8 liter water from pool 1
- fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them
- fill pools 1+2 with exactly 4 liters
- attack Terran with 4-pool