Place the numbers 1 to 64 in the grid, so that each number is a knight's move away from its neighbour. The numbers 1 and 64 are also a knight's move away from each other. So, for example, the numbers 7 and 9 would each be a knight's move away from number 8.

The letters c and s to the right and below the grid indicate where the cubes and squares are situated.

The following clues will help you place the numbers.

C. Only one number in the twenties.
D. Nothing divisible by 21.
F. Only two numbers in the thirties. Total 210.
G. Three numbers in the thirties.

2. Nothing less than 20.
3. No number in the thirties.
4. Total 214.
6. Nothing divisible by 6 or 11. Only one number in the thirties.
7. Only one number divisible by 10.
8. Only one number divisible by 10.

1

2

3

4

5

6

7

8

A

40

54

c

s

s

B

c

C

6

s

D

E

c

s

F

1

14

c

s

s

G

37

59

s

H

19

34

s

s

c

s

c

c

s

s

s

s

c

s

s

AKMA = a knight’s move away
TIOOECA = there is only one empty cell available

From 1-64 the squares are: 1,4,9,16,25,36,49,64
The cubes are: 1,8,27,64

Notice that since a knight’s move shifts you three squares from where you were, each move will always alternate you on a pink/green cell. And since you are moving by consecutive numbers, this means you will be alternating odd/even numbers. So all odd numbers will be on same colored squares, and all even numbers will be on the other same colored squares. Looking at what has been given, we can see that odd numbers will always be on a pink cell and even numbers will always be on a green cell.

Since 1 and 64 are both squares and cubes, whatever cell they are located in must have a C AND an S indicated in both directions. 1 is already placed, so that means one C and one S indicated for column 3 has been claimed (same goes for row F). The only other cells that have an unclaimed C and S indicated in both directions are E3 and E5. 64 must be AKMA from 1, so E5 is 64.

The remaining cubes, 8 and 27, are left to go in A6 and B6. Since 8 is even and 27 is odd, and B6 is green and A6 is pink, this means B6 is 8 and A6 is 27.

In order for 35 to be AKMA from 34, it can only be in F7 or G6. However, 36 will need to be AKMA from both 37 and 35. If 35 is in F7 this will be impossible. So G6 is 35. And since 33 must also be AKMA from 34 and TIOOECA, F7 is 33.

In order for 39 to be AKMA from 40, it can only be in B3 or C2. However, 38 will need to be AKMA from both 37 and 39. If 39 is in B3 this will be impossible. So C2 is 39. And since 41 must also be AKMA from 40 and TIOOECA, B3 is 41.

In order for 38 to be AKMA from both 39 and 37, it can only be in E1 or E3. The clue for column 3 tells us it cannot have any numbers in the 30s. So E1 is 38.

Since 18 and 20 have to be AKMA from 19, they can only be in F2 and G3. The clue for column 2 tells us it has no numbers less than 20. So F2 is 20 and G3 is 18.

Consider cell A8. Since there are only two cells that are AKMA from it (B6 and C7), that means those two cells MUST be A8-1 and A8+1. We already know that B6 is 8, so that means 8=A8-1 or 8=A8+1. This means A8=9 or A8=7. But A8 can’t be 7 since that would mean C7 would have to be 6, which is already placed. So A8 is 9 and C7 is 10.

Row G must have a square number in it. Of the open cells (G1, G5, G7, and G8), only G8 also has an unclaimed S indicated in the column. Therefore G8 is a square number. Since G8 is a pink cell, it must be an odd number. Therefore G8 can only be 25 or 49. Since 26 must be AKMA from both 27 and 25, it is impossible for G8 to be 25. So G8 is 49.

In order for 50 to be AKMA from 49, it can only be in E7 or H6. The clue for column 7 tells us it can only have one number divisible by 10, and there is already one in that column. So H6 is 50. And since 48 must also be AKMA from 49 and TIOOECA, E7 is 48.

51 must be AKMA from 50 and TIOOECA. So F5 is 51.

There are only two cells that are AKMA from 27: B4 and B8. If B8 were 26 then 25 would have to be in C6 or D7. But neither of those cells has an S indicated for both directions. So B4 is 26 and B8 is 28.

Of the cells that are AKMA from 26, only A2 has an S indicated for both directions. So A2 is 25.

Now there are only three square numbers that have not been placed yet: 4, 16 and 36. The only remaining cells that have an unclaimed S in both directions are C3, C4, F4, H3 and H4. Since all three of these square numbers are even, they can only be placed on green cells. So that means they can only go on cells C3, F4 and H4. Since we have limited these three numbers to three cells, we know that C3, F4 and H4 can only be 4, 16 and 36.

All of the cells that are AKMA from H4 are filled: 1, 35, 37, and 51. Since 3 and 5 are not in the set, H4 cannot be 4. And since 15 and 17 are not in this set, H4 cannot be 16. But 35 and 37 are in this set so H4 is 36.

Let’s look at row F. It has three empty cells: F1, F4 and F8. The sum of the existing cells is 119. We know F4 can only be 4 or 16 (neither of which is in the thirties). And we know from the clue for row F that the entire row adds to 210. This means F1+F8 = 87 or 75. The clue for row F also says that it has two numbers in the thirties, and since we only have one number in the thirties so far, we know that one of F1 and F8 are in the thirties. Only 30, 31 and 32 are left. If one of them were 32, which is even, it would have to be on F8 since it is green. But then 32 and 33 would not be AMKA from each other. If one of them were 31, which is odd, it would have to be on F1 since it is pink. But then it would be impossible to place 32 such that it would be AKMA from both 31 and 33. Therefore one of them must be 30, which is even, so F8 is 30 since it is pink. And F1 can only be 57 or 45.

There is only one cell that is AKMA from both 28 and 30, so D7 is 29.

31 must be AKMA from 30 and there are only two empty cells available: E6 and H7. The clue for column 6 tells us it can only have one number in the thirties, and there is already one in that column. So H7 is 31.

32 must be AKMA from 31 and TIOOECA. So G5 is 32.

H2 is only AKMA from three cells. F1 is empty and the others are 1 and 58. Since there is no number that is consecutive to both 1 and 58, we know that H2 must at least be consecutive with F1. We already determined F1 could only be 57 or 45. So H2 can only be 56, 58, 44 or 46. H2 must also be consecutive with 1 or 59. The only number that works is 58. So H2 is 58, F1 is 57, F4 is 4 and C3 is 16.

There are only two cells that are AKMA from both 16 and 18: E2 and E4. The clue for column 2 tells us it has no numbers less than 20. So E4 is 17.

TIOOECA that is AKMA from both 14 and 16, so D5 is 15.
60 must be AKMA from 59 and TIOOECA. So E3 is 60.
56 must be AKMA from 57 and TIOOECA. So D2 is 56.
24 must be AKMA from 25 and TIOOECA. So C1 is 24.
TIOOECA that is AKMA from both 6 and 8, so A4 is 7.

42 must be AKMA from 41 and there are only two empty cells available: D4 and A5. The clue for row D tells us it cannot have any multiples of 21. So A4 is 42.

21 must be AKMA from 20 and there are only two empty cells available: D1, D3 and H3. The clue for row D tells us it cannot have any multiples of 21. So H3 is 21.

22 must be AKMA from 21 and TIOOECA. So G1 is 22.
23 must be AKMA from 22 and TIOOECA. So E2 is 23.
2 must be AKMA from 1 and TIOOECA. So D4 is 2.
TIOOECA that is AKMA from both 2 and 4, so E6 is 3.
TIOOECA that is AKMA from both 4 and 6, so D3 is 5.

The clue for column 4 tells us it adds up to 214. We already have seven of the eight numbers placed, so C4 is 63.

61 must be AKMA from 60 and TIOOECA. So D1 is 61.
62 must be AKMA from 61 and TIOOECA. So B2 is 62.
55 must be AKMA from 56 and TIOOECA. So B1 is 55.
53 must be AKMA from 54 and TIOOECA. So B5 is 53.
TIOOECA that is AKMA from both 51 and 53, so D6 is 52.
11 must be AKMA from 10 and TIOOECA. So E8 is 11.
12 must be AKMA from 11 and TIOOECA. So G7 is 12.
13 must be AKMA from 12 and TIOOECA. So H5 is 13.

Let’s look at C8. It is AKMA from 8, 52, 48 and an empty cell. C8 must be consecutive with at least one of 8, 52 and 48. Since 7, 9, 51, 53 and 49 are already placed, this leaves only 47 as a possibility. So C8 is 47.

46 must be AKMA from 47 and TIOOECA. So A7 is 46.
45 must be AKMA from 46 and TIOOECA. So C5 is 45.
44 must be AKMA from 45 and TIOOECA. So D8 is 44.
TIOOECA that is AKMA from both 42 and 44 (and indeed left on the board). So B7 is 43.

Posted by nikki on 2008-01-09 13:47:39