Professor Stewart's Hoard of Mathematical Treasures (51 page)

Move these two.

The deduction is false.

Suppose, for the sake of argument, that elephants are easy to swallow. Then the third statement in the puzzle tells us that elephants eat honey. The second then tells us that elephants can play the bagpipes. On the other hand, the first statement tells us that elephants wear pink trousers, in which case the fourth statement tells
us that elephants can’t play the bagpipes. So we get a logical contradiction. The only way out is if elephants are not easy to swallow.

There’s a systematic method for answering such questions. First, turn everything into symbols. Let

We use the logical symbols

Then the first four statements read:

We need two of the mathematical laws of logic:

Using these, we can rewrite the conditions as:

so E ⇒ ¬ S. That is, elephants are not easy to swallow.

This list of attributes suggests yet another way to get the answer: think about an elephant (E) that (P) wears pink trousers, (¬ B) does not play the bagpipes, (¬ H) does not eat honey, and (¬ S) is not easy to swallow. Then all four statemnts in the puzzle are true, but ‘elephants are easy to swallow’ is false.

These or their rotations and reflections.

The explanation of Whodunni’s calculator trick uses a bit of algebra.

Suppose you live in house number x, were born in year y, and have had z birthdays so far this year, which is either 0 or 1, depending on dates. Then successive steps in the trick go like this:

If we’re doing the trick in 2009, then 2008 -
y
is one less than the number of years that have passed since your birth year. Adding the number of birthdays you’ve had this year leaves it that way if you haven’t had one yet, but adds 1 if you have. The result is always your age. (Think about it. If you were born one year ago but haven’t had a birthday yet, your age is 0. After your birthday, it’s 1.)

So the final result is 100
x
+ your age. So provided you are aged
between 1 and 99, the last two digits will be your age (written as 01- 09 if your age is 1-9). Removing those and dividing by 100, which is the same as looking at the remaining digits, gives x - your house number.

If you’re over 99, then the final two digits can’t be your age. There will be an extra digit (which barring medical miracles will be 1). So your age will be 1 followed by the final two digits. And your house number will be the rest of the digits except those two, minus one.

If you’re age 0, the trick still works provided you count the day of your birth as a birth-day. Your zeroth, in fact. But usually we don’t do that, which is why I excluded age 0.

To modify the trick for any other year, say 2009 +
a
, just change the final step to ‘subtract 42 - a’. So in 2010 subtract 41, in 2011 subtract 40, and so on. If you’re reading this after 2051, make that ‘add
a
- 42’. It’s the same thing, but it will sound more sensible.

To subtract (say) a 3-digit number [x][y][z], which is really 100
x
+ 10
y
+
z
, we have to form the complement [10 -
x
][10 -
y
] [10 -
z
], which is really 100(10 -
x
) + 10(10 -
y
) + (10 -
z
). This is equal to 1000 - 100
x
+ 100 - 10
y
+ 10 -
z
, or 1110 - (100
x
+ 10
y
+
z
). So adding the complement is the same as subtracting the original number, but adding 1110. To get rid of it, subtract 1 from positions 4, 3, 2, but not 1.

Redbeard will locate the lost loot 128 paces north of the rock.

At each step the piratical finger can move either left or right - two choices. So the number of routes down the diagram doubles for each extra row. There are 8 rows, and only one T to start from, so the number of routes is 1×2×2×2×2×2×2×2 = 128.

If we replace each letter by the number of routes that lead to it, we get a famous mathematical gadget, Pascal’s triangle:

Here each number is the sum of the two above it to left and right, except down the sides where they’re all 1’s. If you add up the rows, you get the powers of two: 1, 2, 4, 8, 16, 32, 64, 128. So this is another - closely related - way to get the same answer.

Fold the paper in half (say along the vertical line in my picture) and then fold it alternately up and down along the other lines to make a zigzag shape, the way a fan folds. Then cut along a suitable slanting line - and unfold. I’ve drawn the ghost of the star to show how it relates to the folds.

Fold . . .

... and cut.

Yes, you can make a six-pointed star in a similar way. If anything, it’s easier: first fold the paper in quarters, then fold the result along two lines trisecting the right-angled corner. As for the five-pointed star, you have to snip at the correct angle. That’s committees for you.

The cycles that appear with zero or negative numbers are:

The lengths contain 8, 7, 6, 6, 5, 5, 5, 4 and 3 links. Instead of breaking up one link on each chain, we could break all 8 links in the 8-link piece, and use these to join the other eight pieces together: total cost £24. But there’s a cheaper way. Break the pieces of lengths 4 and 3 into separate links, and use these to join the seven other pieces. The total cost is now £21.

No, it doesn’t wave its paws madly and exploit air resistance to create a force, like the wings of a bird do. Instead, the cat manages to change its orientation without causing any change to its angular momentum at any time.

No contradiction there, but of course there’s the bit in between, when the cat starts to rotate. Except - it doesn’t. Rotate, that is. A cat is not a rigid body.
⁵²
In 1894, the French doctor Étienne Jules Marey took a series of photographs of a falling cat.

Marey’s cat experiment.

The secret was then revealed. Because a cat is not a rigid body, it does not have to rotate its entire body simultaneously. Here is the cat’s recipe for turning over, while maintaining zero angular momentum throughout: