- 08 Oct '06 13:45Try these:

1. 100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world can always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.

Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion?

2. We know that the derivative of x2 with respect to x is 2x. However, what if we rewrite x2 as the sum of x x's, and then take the derivative:

d/dx[ x2 ] = d/dx[ x + x + x + ... (x times) ]

= d/dx[x] + d/dx[x] + d/dx[x] ... (x times)

= 1 + 1 + 1 + ... (x times)

= x

This argument shows that the derivative of x2 with respect to x is actually x. So what's going on here?

3. There are two envelopes in front of you each with a non-zero sum of money. You are informed one has twice as much money as the other. You are then allowed to select either envelope and keep the money inside. After you select one and before opening it you are given the option to change your mind and switch to the other one? You think to yourself that if your envelope has x dollars there is a 50% chance the other one has x/2 dollars and a 50% chance it has 2x dollars. The expected return, you compute, is .5[.5x + 2x]=1.25x which seems like a favorable gamble. Do you switch and why? Assume you are neither risk averse nor risk prone, in other words you will take any good gamble and avoid any bad one.

4. The professor for class Logic 315 says on Friday: "We're going to have a surprise quiz next week, but I'm not telling you what day... if you can figure out what day it will be on, I'll cancel the quiz."

The students get together and decide that the quiz can't be on Friday, as if the quiz doesn't happen by Thursday, it'll be obvious the quiz is on Friday. Similarly, the quiz can't be on Thursday, because we know it won't be on Friday, and if the quiz doesn't happen by Wednesday, it'll be obvious it's on Thursday (because it can't be on Friday). Same thing for Wednesday, Tuesday and Monday. So it can't be on ANY day, so there's no quiz next week!"

They tell the professor, who smiles and says, "Well, nice to see you're thinking about it."

On Tuesday, the professor gives the quiz, totally unexpected!

What's the flaw in the students' thinking?

5. Below are 10 statements, all either true or false:

1. At least one of statements 9 and 10 is true.

2. This is either the first true statement or the first false stament.

3. There are three consecutive false statements.

4. The difference between the number of the last true statement and the first true statement divides the number which is to be found.

5. The sum of the numbers of the true statements is the number which is to be found.

6. This is not the last true statement.

7. The number of each true statement divides the number which is to be found.

8. The number that is to be found is the percentage of true statements.

9. The number of divisors of the number that is to be found (apart from 1 and itself) is greater than the sum of the numbers of the true statements.

10. There are no three consecutive true statements.

Find the minimum admissible number (which is to be found). - 09 Oct '06 16:35 / 1 edit1.

The prisoner that is called on the first day is going to be the one to make the claim (lets call him the Counter)

Counter's actions: If light is off -> turn light on

If light is on -> leave light on

All other prisoners:

Prisoner who has turned light off once will never turn it off again. Prisoner who has never turned the light off will turn the light on if it's on. They never turn the light on.

This'll take a long time, but if the Counter has seen the light off for 100 times he can immedeately claim that all prisoners have been in the room (99 times the prisoners have turned it off plus the very first day the Counter entered the room).

Now they just need a guard who will make sure all prisoners get out enough times - 09 Oct '06 19:151. This is the answer i have but i am assured by others that there is a quicker way

2. Yes i think u understand this one - basically x is a variable so the first line does not make sense in a differential

3. Yes but explain why the argument in the question is false

4. Yes, this one took me a while to get my head around. Easy to understand the answer difficult to uderstand the fallacy in the question

5. ?