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

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Authors: Ian Stewart

Tags: #Mathematics, #General

BOOK: Professor Stewart's Hoard of Mathematical Treasures
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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 Look and Say Sequence
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
L(n)
≈(1.30357726903...)
n
where 1.30357726903 . . . is the smallest real solution of the 71stdegree polynomial equation
x
71
—
x
69
—2x
68
—x
67
+ 2x
66
+ 2x
65
—x
63
—x
62
—x
61
—x
60
+ 2x
58
+ 5x
57
+ 3x
56
—2x
55
—10x
54
—3x
53
—2x
52
+ 6x
51
+ 6x
50
+ x
49
+ 9x
48
—3x
47
—7x
46
—8x
45
—8x
44
+ 10x
43
+ 6x
42
+ 8x
41
—5x
40
—12x
39
+ 7x
38
—7x
37
+ 7x
36
+ x
35
—3x
34
+ 10x
33
+ x
32
—6x
31
—2x
30
—10x
29
—3x
28
+ 2x
27
+ 9x
26
—3x
25
+ 14x
24
—8x
23
—7x
21
+ 9x
20
+ 3x
19
—4x
18
—
10x
17
— 7x
16
+ 12x
15
+ 7x
14
+ 2x
13
— 12x
12
— 4x
11
— 2x
10
+
5x
9
+
x
7
—
7x
6
+
7x
5
—
4x
4
+
12x
3
—
6x
2
+
3x
—
6
= 0
I don’t claim this is obvious.
The Millionth Digit
The millionth digit is 1.
The numbers 1-9 occupy the first 9 positions.
The numbers 10-99 occupy the next 2 × 90 = 180 positions.
The numbers 100-999 occupy the next 3 × 900 = 2,700 positions.
The numbers 1,000-9,999 occupy the next 4 × 9,000 = 36,000
positions.
The numbers 10,000-99,999 occupy the next
5 × 90,000 = 450,000 positions.
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.
Piratical Pathways
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.
• Start by writing 1 next to the P.
• There is exactly one path from the P to each of the I’s, so we write 1 next to each I.
• Look at each R in turn, see which I’s connect to it, and add up the numbers beside those letters. Here one of the R’s is connected to only one I, numbered 1, so it also gets the number 1. The other three are connected to two I’s, numbered 1, so they are given the number 1 + 1 = 2.
• Next, move on to the A’s. The leftmost A is connected to three R’s: one numbered 1 and two numbered 2, so we give that A the number 1 + 2 + 2 = 5. And so on.
• Continuing in this way, we eventually reach the final E. The T’s that connect to it are numbered 7, 7 and 5. So we number E with 7 + 7 + 5 = 19. And that’s the number of ways to get to E.
Trains That Pass in the Siding
Yes, they can pass each other - however long the trains might be.
Here’s how.
1. Initially, each train is on its side of the siding.
2. Train B backs away to the right. Train A runs right, past the siding, backs on to it, drops off four coaches, returns to the main line going right, and backs off to the far left.
3. Train A moves past the siding, going left, and joins the main part of train B.
4. Trains A+B move right, going along the siding, pick up the four coaches, and return to the main line on the right of the siding.
5. Then they back up on to the siding, drop off four more coaches, and return to the main line on the right of the siding.
6. The main part of the combined trains A+B moves left along the main line until it clears the siding.
7. Again trains A+B move right, going along the siding, pick up the four coaches, and return to the main line on the right of the siding.
8. Then they back up on to the siding, drop off one coach and locomotive A, and return to the main line on the right of the siding.
9. The main part of the combined trains A+B moves left along the main line until it clears the siding.
10. Finally, A+B goes right along the siding, rejoins with locomotive A and its coach. Then the trains split apart and each continues on its way.
The same method works no matter how long the trains are, provided the siding can contain at least one coach or locomotive.
Squares, Lists and Digital Sums
The next such sequence is
99,980,001, 100,000,000, 100,020,001, 100,040,004 100,060,009, 100,080,016, 100,100,025
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,
16
2
= 256 with digit-sum 13
19
2
= 361 with digit-sum 10
(20
2
, 21
2
, and 22
2
have square digit-sums so I can’t use those)
25
2
= 625 with digit-sum 13
29
2
= 841 with digit-sum 13
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.
Match Trick
It’s easy if you let the edges of the triangles overlap.
Slicing the Cake
You can get at most 16 pieces. Here’s one way to do it:
How to get 16 pieces with five cuts.

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