Professor Stewart's Hoard of Mathematical Treasures (62 page)

The stated conditions do not require you to use the integers 1-9, and in fact no solution exists if you do, because then the even numbers must go in the corners. But by using fractions, you can solve the puzzle. The figure shows the traditional solution, perhaps the
simplest, but there are infinitely many others, even if you restrict the entries to positive numbers.

An unorthodox magic square.

The rule of formation is best stated in words. The first term is ‘1’, which can be read as ‘one one’, so the next term is 11. This reads ‘two ones’, leading to 21. Read this as ‘one two, one one’ and you see where 1211 comes from, and so on.

Conway proved that if
L
(
n
) is the length of the nth term in this sequence, then
where 1.30357726903 . . . is the smallest real solution of the 71stdegree polynomial equation
I don’t claim this is obvious.

The millionth digit is 1.

At this stage, we have reached the 488,889th digit altogether. Since 1,000,000 — 488,889 = 511,111, we want the digit in the 511,111th place, in the block that starts 100,000-100,001-100,002- and so on. Since these are grouped in sixes, we work out 511,111/6 = 85,185
. Therefore we are seeking the first digit of the 85,186th 6-digit block. That block must be 185,185, and its first digit is 1.

Redbeard’s bank is at 19 Taxhaven Street.

Calculating the number of paths.

The number is small enough that you can just list the possible paths, but there’s a systematic method for this sort of question. The diagram shows the same map; I’ve removed superfluous connections that can never be used to keep thing simple, but it makes no difference to the method or the result if you leave them in.

I’ve written numbers beside the letters. These numbers tell us how many ways there are to reach that particular letter, and we calculate them in turn: P, the three I’s, the four R’s, the three A’s, the three T’s, and the final E.

Yes, they can pass each other - however long the trains might be.

The same method works no matter how long the trains are, provided the siding can contain at least one coach or locomotive.

The next such sequence is
which are the squares of the numbers 9,999-10,005.

A good place to look is the squares of numbers 100 ... 00,
100 . . . 01, 100 . . . 02, 100 . . . 03, 100 . . . 04, 100 . . . 05, which have lots of zeros, while the few remaining digits give digital sums that are the squares 1, 4, 9, 16, 16, 9. To extend this list of six consecutive squares to seven, we have to look at 999 . . . 9 and 100 . . . 06. The digits of 99
²
sum to 18, not a square; those of 999
²
sum to 27, also not a square. But the digits of 9999
²
sum to 36, a square. Looking at the other end, the digits of 106
²
, 1,006
²
, and 10,006
²
sum to 13, not a square.

To rule out anything in between 15
²
and 9999
²
, we just have to find a sequence of squares of numbers that differ by at most 6, whose digits sum to non-squares. For example,
and so on. I’m sure there must be short cuts, and a computer can quickly check all possibilities in that range.

No one seems to know whether eight consecutive squares can all have square digit-sums.

It’s easy if you let the edges of the triangles overlap.

You can get at most 16 pieces. Here’s one way to do it: