Professor Stewart's Hoard of Mathematical Treasures (64 page)

The analysis depends on prime numbers, which divide into several kinds: big primes greater than 100/2 (53, 59, 61, 67, 71, 73, 79, 83, 89, 97), medium large primes between 100/2 and 100/3 (37, 41, 43, 47), medium primes between 100/3 and 100/4 (29, 31), small primes less than 100/4 but not too small (17, 19), and very small
primes (2, 3, 5, 7, 11). The winning openings are twice the medium primes. For example, here is the analysis if Mathophila opens with 58.

In his number theory course, Wigner solved the whole thing by providing a criterion for winning play in all cases. The answer for JG-n depends on whether the various powers of primes that occur in the factorisation of n! are odd or even.

Slade’s braid trick relies on a topological curiosity: his leather strip can be deformed, in perfectly ordinary 3D space, to end up braided. So all he had to do was fiddle with it under the table until it reached the braided state. I’ve separated the three strips in my picture to make the method clearer.

This sequence of moves introduces six extra crossings of the three strips, so by repeating it you can make really long braids.

 

Slade had a colourful career, and was exposed as a fraud by the Seybert Commission in 1885. See:
www.answers.com/topic/henry-slade

How to keep the neighbours apart.

This is the only solution, aside from rotations and reflections.

The point on the rim of the wheel, where it touches the road, has instantaneous velocity zero. The ‘no slip’ condition means that the
horizontal component of velocity at this point is 0; the ‘no bounce’ condition means that the vertical component here is also 0.

This is interesting, because the point concerned moves along the road at 10 metres per second. But as it moves, the point on the road corresponds to different points on the wheel. And the question was about points on the wheel, not points on the road.

A more detailed analysis using calculus shows that this is the only such point. Assume the wheel starts with its centre at (0, 1) and rolls along the x-axis to the right. Place a black dot on the rim, starting at the origin (0, 0) at time 0.

After time t the circle has rolled 10t to the right, and has also turned clockwise through an angle 10t. So the black dot will now be at the point
Its velocity vector is the derivative of this with respect to t, which is
This vanishes when
That is, 10t = 2nπ for integer n, or t = nπ/5. But at these times the dot is at positions (2
n
π, 0), which are the successive points at which the dot hits the ground.

The same kind of calculation shows that any point not on the rim always has non-zero velocity. I’ll omit the details.

It can be proved - not easily - that the process cannot continue past the 17th point.

The first proof was found by Mieczyslaw Warmus, but this wasn’t published; the first published proof was given by Elwyn Berlekamp and Ron Graham in 1970. Warmus then published a simpler proof in
1976. He also proved that there are precisely 1,536 distinct patterns for placing 17 points, which form 768 mirror-image pairs.

White can force a win by moving the knight.

This is the only opening that can force a win, but I’ll omit that part of the analysis.

To see why moving the knight leads to a win, number the cells of the board 1-8 from the left. Use the symbols R = rook, N = knight, K = king, × = takes, - = moves, * = check, † = checkmate. The table shows only some of the possible sequences of moves, namely those in which White makes one move (which eventually leads to a win whatever Black does) at each step. All Black’s possible replies are considered. This technique is called ‘pruning the game tree’, and it works provided White wins for every line of play that is included. What it omits are alternative ways for White to win, if they exist, and any White moves that could lead to a forced loss for White.

You can’t win. The Infinite Lottery always beats you by forcing you to remove all the balls.

This may seem rather counter-intuitive, given the way the total number of balls can increase by gigantic amounts at each step. But these amounts are finite; infinity isn’t. Raymond Smullyan proved in 1979 that you always lose. His idea is to look at the biggest number in the box, and keep track of the balls that bear that number.

First, suppose that the biggest number in the box is 1. Then all
balls bear the number 1. So you have to remove all the balls, one at a time - which means you lose.

Now suppose that the biggest number in the box is 2. You can’t keep discarding 1’s indefinitely, because they will eventually run out. So at some stage you have to discard one of the 2’s and replace it with lots of 1’s. Now the number of 2’s has decreased. The number of 1’s has gone up, but it’s still finite. Again, you can’t keep discarding 1’s indefinitely, so eventually you have to discard another of the 2’s and replace it with lots of 1’s. Now the number of 2’s has decreased again. Every so often you have to discard a 2, so eventually you run out of 2’s altogether. But now all the balls in the box are 1’s - and we’ve already seen that in that case you lose, however many 1’s there may be.

Ah, but maybe the biggest number in the box is 3. Well ... you can’t keep choosing (and discarding) 2’s and 1’s for ever, for the reasons we’ve just discussed. So eventually you have to discard a 3. Now the number of 3’s drops by one, and the same argument shows that you have to discard another 3 at some point, and another, until you run out of 3’s. Now the box contains only 1’s and 2’s, and we’ve just seen that in this case you lose.

Continuing in this way, it’s clear that you lose if the biggest number in the box is 4, 5, 6, . . . , and so on. That is, you lose no matter what the biggest number in the box is. But the number of balls in the box is finite, so there must be some biggest number.

Whatever it is, you lose!

Formally, this is a proof by the Principle of Mathematical Induction. This principle states that if some property of whole numbers n holds for
n
= 1, and its truth for any n implies its truth for n + 1, then it holds for all whole numbers. Here the property concerned is ‘If the biggest number in the box is n, then you lose.’

Let’s check that. If
n
= 1, then the biggest number in the box is 1, and you lose.

Now, suppose that we have proved that if the biggest number in the box is n, then you lose. Suppose that the biggest number in the box is n + 1. You can’t keep discarding numbers n or less, because we know that if you do, you lose - that is, you run out of balls numbered n or less. So at some point you must discard one of the balls bearing
the number n + 1, and the number of such balls drops by one. For the same reason, that number must drop again, and again ... and eventually you discard all the balls marked n + 1. But now the remaining balls bear numbers n or smaller, so you lose. In short, if the biggest number in the box is n + 1, then you lose. And that’s the other step you need to complete the induction proof.

You can make the game go on for as long as you like, but it must stop after finitely many moves. However, that finite number can be as big as you wish.

13 ships.

Suppose (the date doesn’t matter, but this choice makes the sums simpler) that the New York ship sets sail on 10 January. It arrives on 17 January, just as the 17 January ship from London departs.

Similarly, the ship that left London on 3 January arrives in New York on 10 January, just as the ship we’re talking about leaves.

So on the high seas, our ship encounters the ones that left London starting on 4 January and ending on 16 January. That’s 13 ships in all.

The complicated calculation is pure misdirection. The fallacy is the assumption that such a number n exists. This illustrates a key aspect of mathematical proofs: if you define something by requiring it to possess some particular property, you can’t assume that ‘it’ has that property unless ‘it’ exists.

In this case, it doesn’t.