MrRubix's Riddle/Problem Thread: Rebirth

[MrRubix]

Originally Posted by ieatyourlvllol

Hmm...the pole vault method would probably be the most suitable. Additionally, to reduce the possibility of uncontrolled spillage upon catching the cup, you could try fashioning a net out of your clothing. Had there been an additional restriction like "you're not allowed to put the cup on the ground," I would have suggested that if your window was particularly big, you could make further use of your garments by wrapping them around the cup to temporarily inhibit seepage on a subsequent throw (out the window). Otherwise (if the risk was too high), you would use a similar process, except that the bundle would be tied around the pole with the cup base perpendicular. In this case, the gasoline would first contact the fabric at roughly halfway through your vault, but if possible (assuming average absorption and permeability), leakage would take longer than what would be required for you to finish your escape.

Right on

[Fractal_Monkey]

No-one else going for the math one? Okay then, I'll try it.

Evaluate (Σ of k=2 to infinity) for (ζ(k)-1)
Where the zeta function (ζ) is defined as:
ζ(s) = (Σ of n=1 to infinity) for 1/(n^s)

Given that ζ(s) - 1 = 1/2^s + 1/3^s + 1/4^s + ...

(Σ of k=2 to infinity) for (ζ(k)-1) can be expressed as:

(1/2^2 + 1/3^2 + 1/4^2 + ...) + (1/2^3 + 1/3^3 + 1/4^3 + ...) + (1/2^4 + 1/3^4 + 1/4^4 + ...) + ...

These terms can be re-ordered as:

(1/2^2 + 1/2^3 + 1/2^4 + ...) + (1/3^2 + 1/3^3 + 1/3^4 + ...) + (1/4^2 + 1/4^3 + 1/4^4 + ...) + ...

Each bracketed series is a geometric progression with initial term 1/n^2 and ratio 1/n (n = 2, 3, 4, ...), so the sum of the general bracket is (1/n^2)(n/n-1) = 1/n(n-1). Therefore:

(Σ of k=2 to infinity) for (ζ(k)-1) = (Σ of n=2 to infinity) for (1/n(n-1))
= (Σ of n=1 to infinity) for (1/n(n+1))

Now, 1/n(n+1) = 1/n - 1/(n+1) , so

(Σ of n=1 to infinity) for (1/n(n+1)) = (Σ of n=1 to infinity) for (1/n) - (Σ of n=1 to infinity) for (1/(n+1))
= (1 + 1/2 + 1/3 + 1/4 + ...) - (1/2 + 1/3 + 1/4 + 1/5 + ...)
= 1 + 1/2 - 1/2 + 1/3 - 1/3 + 1/4 - 1/4 + ...
= 1 + 0 + 0 + 0 + ...
= 1

[MrRubix]

Correct

I'll post new riddles later

[emerald000]

I felt like posting a mathematical riddle of my own. It's a relatively simple problem within the reach of anyone. No need to use calculus or anything of the sort.

Assuming a > b > 0, order the harmonic mean, the geometric mean, the arithmetic mean and the root mean square in a relation of inequality. (Exemple: HM < AM < RM < GM)

Harmonic mean: 2ab/(a+b)
Geometric mean: sqrt(ab)
Arithmetic mean: (a+b)/2
Root mean square: sqrt((a²+b²)/2)

Ordering the inequalities is easy, proving it is somewhat harder.

[MrRubix]

If nobody else is gonna take that one, I will, lol. I'll give someone else a chance though.


I'll add some more stuff later tonight.

[Fractal_Monkey]

I've got the ordering and a proof, but I answered the last one. It's someone else's turn.

[orlando2k7]

I was close to the answer for the math one.

I had it after the answer was revealed. :/

[MrRubix]

New riddles up

[Bolth mannn]

i know its already been answered...but id just control-spill the cup directly in the middle of the room and make sure it doesnt spread to the fire then just WAIT for help

EDIT: is the room big enough for me to do that?

[FuzenAkuma]

Wouldn't the probability of choosing the two-headed quarter just be 1/573?

[emerald000]

Yeah, the first answer is 1/573.

But for the second one, do we suppose the rubbing alcohol is 100% ethanol, or something like 90% ethanol-10% water? Assuming it is 100% pure, the final concentration is given as C=1/(1+x) where there is 1 unit of concentration at the start and there is x units removed from the tube.

And for my riddle, you may PM me or AIM me the solution if you don't want to make sure it is OK before posting it.

[ieatyourlvllol]

The answer to the second riddle is one (simplified from a series of bivariable expressions/equations).

I'm pretty sure the Konami riddle is based on dependent probability. I might go ahead and verify the answer later.

[emerald000]

Clearer answer for #2:

If we put the water in the alcohol, we get, at the end, a concentration of 1/(1+x) of water in the water tube and 1/(1+x) of alcohol in the alcohol tube.

EDIT: Did the coin flip mathematically and it gave 49/621 or around 7.89%.

[ieatyourlvllol]

My calculation (using pure alcohol) was as follows:

{(v-x+[(x*[x/(v+x)])/v])/[(v*[v/(v+x)])/v]} = 1(:1 ratio)

That value would probably change (and the system would gain another variable) if the rubbing alcohol had a partial concentration.

[MrRubix]

The answer to the first one is not merely 1/573. It is a conditional probability. You know you flipped a certain number of heads. You don't, however, know where they came from.

As for the alcohol problem, if you're using a bunch of algebra, you're working too hard. You are right though that the final concentrations are identical.

When amount X is poured, the final concentration of alcohol in the alcohol tub is V/(V+X). After mixing and pouring back, the concentration of alcohol in the alcohol jug does not change again because no water is added. However, when the diluted alcohol is poured back into the water tub, the concentration of water in the water tub changes from 100% to V/(V+X). So, again, the final concentrations are the same.

At the end of the process, both tubs contain the same volume of fluid as they did at the start. The only way for the concentration of alcohol to have changed from 100% is if some alcohol was displaced by water. Vice-versa for water. Volume is conserved (both total volume and volume in each tub), so all that has happened is that identical quantities of water and alcohol have traded places (and these identical quantities are slightly less than X). So, by symmetry, the concentrations of alcohol in the alcohol tub and water in the water tub must be identical.

[ieatyourlvllol]

Yeah, algebra really wasn't necessary. I just like doing things mathematically. 8)

The simplest method would be to visualize extremes. You could just as well say that the amount 'X' that you pour from Tub 1 is equal to either 0 or the volume of liquid. In both cases, it's easy to see that the concentrations will end up the same. As a bonus, any intermediary value of 'X' would have to abide by the same laws of proportional equivalence.

I wanna take a crack at the Konami riddle, but I'm too busy right now. =(

[emerald000]

Originally Posted by emerald000

EDIT: Did the coin flip mathematically and it gave 49/621 or around 7.89%.

Is it the right answer?

And why is there a volume variable in your equations? Concentrations are going to be the same no matter what is the initial volume. If you have twice the water, you'll have twice the alcohol, thus simplifying themselves...

[MrRubix]

No, 7.89% is not correct

"The simplest method would be to visualize extremes. You could just as well say that the amount 'X' that you pour from Tub 1 is equal to either 0 or the volume of liquid. In both cases, it's easy to see that the concentrations will end up the same. As a bonus, any intermediary value of 'X' would have to abide by the same laws of proportional equivalence."

That's also a very good way to look at it -- nice job

emerald: It's there for ease of understanding. You can solve the riddle without any math -- just understanding the underlying concept is sufficient to solve it.

[dooty_7]

here is a stab in the dark for the coin flip one...although I will end up sounding retarded :P

The probability of flipping seven heads in a row on a standard coin is 1 / 2^7 (or just 0.5^7) and for the two headed coin the probability is obviously 1.

So the probability that the coin you picked is the 2 headed coin is
(572/573) * (1/2^7) + (1/573)*(1) = 0.009544 which is greater then just 1/573

EDIT: On second thought, I think this solution just gives the probability of getting 7 heads in a row lol

[MrRubix]

If you're going for the Konami riddle, you need to approach it conditionally. There is a very well-known statistical/probability tool for solving these kinds of problems. You KNOW you already got 7 heads in a row. Therefore you aren't solving for "the probability of getting 7 heads" in the strictest sense. You're being asked "you ALREADY got 7 heads. Did these heads COME FROM a fair coin or the two-headed coin?"


Dooty: Yes, that would be right had I asked "What is the probability of getting 7 heads in a row?"
P(getting 7 heads) =
Splitting this event into mutually exclusive events:
P(getting 7 heads fair) + P(getting 7 heads with the double headed coin) =
P(choosing fair coin)*P(getting 7 heads with fair coin) + P(choosing double headed coin)*P(getting 7 heads with double headed coin) =
(572/573)*(1/2)^7 + (1/573)*1