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Your Favorite Math/Logic Riddles?
shma asks: "Whether you're involved in the Sciences, Mathematics, or Engineering, you undoubtedly enjoy finding simple solutions to seemingly difficult problems. I'm sure you all have a favorite mind-bender, and who better to share it with than the Slashdot community? Post your own problems and try to solve others. Just one request: If you have figured out the solution, link to it in a post, rather than write it out where anyone can see it." What brain benders tickle your fancy?
"Here's a sample to consider: You're in a dark room with 50 quarters, 18 of which are heads up. You are allowed to move around the coins or flip some or all of them, if you wish. Problem is, it's too dark to tell what you're moving or flipping (no, you can't figure it out by touch either). Your job is to split the coins into two groups, each of which has the same number of heads up coins. How do you accomplish this?"
There's a good write up of this on MathWorld.
Uhmm, if n = 0, that is not true.
a^0 = 1
b^0 = 1
c^0 = 1
1 != 2
So, I would submit that that might be true for all nonzero values of n.
Well, If b=0, b^0 is not even defined ! For n=0, it is easy to see that there is no solution. For n smaller than 3, it is elementary to show that there are solutions (even infinitely many of them), and for n > 3, you have to be Andrew Wiles to show that :)
Is ..:: Riddles ::... In has (amogst others) the famous "prison with a lamp" problem:
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?
a = first 3 digits
b = last 4 digits
((a*80+1)*250 + b+b -250)/2
(a*20000 + 250 + b*2 - 250)/2
a*10000 + 125 + b - 125
a*10000 + b
It's only amazing if you don't know algebra, and no, a calculator is not required. Then again, if the point is to encourage people to eventually put down their calculator and instead try understanding why something works, then I'm all for it.
Dividing by zero is not "perfectly valid algebra". Division is not closed on the set of real numbers. Its not really a riddle if you lie in the problem description. Otherwise the solution to the sample problem could be "Pull out 9 of the quarters into a seperate group. I was lying when I said you couldn't see any of them."
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
Weak ;)
(20000a + 250 + 2b - 250) / 2
10000a + b
It's a terribly written version of the "warden and his prisoners" problem, which you can google. The only difference other than confusing language is the original problem has two on/off switches, and each prisoner must flip one, rather than one that can be left alone.
*Spoiler* Don't read the following if you don't wanna know the answer:
1) The prisoners elect one of their own to be a counter, the rest we will call non-counters.
2) When a non-counter comes into the chalice room, if he can he will put the chalice right side up. If it's already right side up, he'll leave it alone. However, each non-counter will only do this once. If he's already flipped it in the past, and it's upside down, he'll leave it upside down.
3) Every time the counter comes in, he checks the chalice. If it's upside down, he'll do nothing. If it's right side up, he'll flip it, and add one to his count. Once he's flipped it n times (n being the prisoner count), he knows everyone has done it. If the original state of the chalice is known, the problem can be modified so he only needs to flip it n-1 times.
Send lawyers, guns, and money!
I had a conversation with Brian0918 on AIM this morning, in which he revealed he's really trolling when I pointed out to him there is no solution (see my other posts) on this topic. Here's a little tidbit: ". i usually just post the problem to get people into big disputes, which so far has worked 2 out of 2 times". If you want the full conversation, email me at sbartaNOSPAM_at_MAPSONgmail.com.
Rank my idea: http://www.sinceslicedbread.com/node/531