142857 x 9 = 1285713 (3 + 1 replaces the missing 4)

142857 x 10 = 1428570

142857 x 11 = 1571427

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it's a drawn game, but black has decided that he'll resign, on one condition: you must tell him the minimum number of moves necessary for white's bishop to travel from h1 to a8, whilst passing through all the other white squares of the board once, and only once. (kings remain stationary)

That's what's so befuddling about the double-or-half paradox. The math seems to add up perfectly but it defies all common sense. I don't have a pat answer.

I have told this to my math-minded friends, who were very upset to hear all of this, and to preserve their sanity they came up with DIFFERENT mathematical models of the situation which did reach the correct conclusion. The correct conclusion, of course, being that it makes no difference if you switch.

For example, forget about my "X", let's just say that the envelope with the smaller amount is Y and other one contains 2Y. Now, whichever one you pick, your expected return is 1.5*Y, and switching will not change this simple fact. Voila, problem solved.

Solved? Not at all. He never showed the flaw in my reasoning, he just provided different reasoning. That doesn't solve the paradox. If anything it shows that mathematics is built on a foundation of sand, since two perfectly acceptable mathematical methods yield entirely different results.

A chain of mass "M" and length "L" is suspended vertically with its lower end just touching a scale. The chain is released suddenly and falls onto the scale. What is the reading of the scale when length "x" of the chain has fallen? Neglect sizes of individual chain links.

I think it's an inexact problem though, and if you want to be accurate, it can actually be fiendishly difficult. I'll briefly outline a few of these points:

To start with, if we assume that the chain is hung perfectly straight, then when it is released, we actually require some sort of asymmetry of the chain for it to fall over into a bundle on the scale. Otherwise, assuming perfect symmetry, there's no reason for it to fall one way or the other, and it will stand on end (until the quantum fluctuations get it, which may not be long - it depends on the shape of the head of the chain).

In this rather interesting case, the bottom of the chain is not going anywhere, while the top starts to fall, so the chain will compress. This compression will continue to increase until it is sufficient to start pushing the top back up again. And you see that it will behave like a spring, expanding and contracting about it's equilibrium forever (or until the energy dissipates, obviously). In this case, we have an oscillatory reading on the scale!

Then there's the possibility of the chain bouncing off the scale when it lands, so I think we also need to assume that the collision is completely inelastic (i.e., the kinetic energy is converted purely into sound and heat, say), otherwise the problem is getting beyond the calculable realm! If the collisions are perfectly elastic (i.e., bouncy chain!), and the chain is much lighter than the earth, we'd probably have a periodic (maybe even chaotic(!), depending on how the chain was precisely arranged when it was dropped) solution, with the chain continually bouncing up and down off the scale.

let's suppose that you're holding a king and a 5. when monty hall (dressed up to look like the devil) decides to make a deal with you.

he offers you two cards face down. the first one he says, is an 8 of clubs, and the other one is a 4 of hearts.

you take one leaving it face down. he then offers you to make the switch.

if we say that the 4 is Y, and the 8 is 2Y, then you're expected return is 1.5*Y, which is 6, and you'll get 21 if you keep switching ad infinitum (and it's a sure bet, you'll go to tahiti).

Let a = 1, b = 1.

a = b <identity theorem>

a^2 = ab <multiply by a>

a^2 - b^2 = ab - b^2 <subtract b^2>

(a-b)(a+b) = b(a-b) <factor>

a+b = b <divide>

1+1 = 1 <substitute>

The reading of the scale consists of two parts: the weight of the chain accumulated and the impulse per unit time imparted by the chain colliding w/ the scale. Let's examine these two parts:

1) The weight accumulated is simply: W = (Mgx)/L ("g" is of course the gravitational constant)

2) Keep in mind that obviously the velocity of the links at the instant of hitting the scale is found from v*v = 2gx. The impulse per unit time is dp/dt. p=mv always, and so we can arrange as follows: d(mv)/dt = v(dm/dt) = v*(M/L)*v. Here, at the last stage, we can substitute from the velocity formula and obtain: 2Mg(x/L)

The solution is to sum parts (1)+(2):

Mg(x/L)+2Mg(x/L) = 3Mg(x/L)

By the way, since the above problem was not very difficult, I can ask a much trickier(and more interesting) problem if anyone would like?