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The puzzles are presented in the order of difficulty. The four last puzzles are open, i.e. only row and column sums are given. Each of them has a unique solution. However, there are often many different ways to find that solution. Although it is permitted to attain the solution by pure guesswork, a more appreciated and demanding way is to be able to show that only one solution exists. All solutions given here fulfill the latter requirement. Please note how different strategies are available.

More information about Survo Puzzles

A | B | C | D | ||
---|---|---|---|---|---|

1 | 2 | 10 | 23 | ||

2 | 9 | 11 | 24 | ||

3 | 12 | 8 | 31 | ||

21 | 15 | 19 | 23 |

A | B | C | D | ||
---|---|---|---|---|---|

1 | 2 | 6 | 33 | ||

2 | 4 | 1 | 26 | ||

3 | 5 | 13 | 42 | ||

4 | 14 | 3 | 35 | ||

38 | 33 | 34 | 31 |

A | B | C | D | ||
---|---|---|---|---|---|

1 | 26 | ||||

2 | 33 | ||||

3 | 19 | ||||

31 | 7 | 25 | 15 |

A | B | C | D | E | F | G | ||
---|---|---|---|---|---|---|---|---|

1 | 77 | |||||||

2 | 28 | |||||||

3 | 126 | |||||||

36 | 39 | 30 | 33 | 42 | 27 | 24 |

A | B | C | D | ||
---|---|---|---|---|---|

1 | 49 | ||||

2 | 23 | ||||

3 | 34 | ||||

4 | 30 | ||||

40 | 52 | 10 | 34 |

A | B | C | D | E | F | G | H | I | J | ||
---|---|---|---|---|---|---|---|---|---|---|---|

1 | 64 | ||||||||||

2 | 146 | ||||||||||

10 | 11 | 14 | 17 | 20 | 22 | 23 | 27 | 31 | 35 |

A B C D 1 * 2 10 * 23 2 9 * * 11 24 3 * 12 * 8 31 21 15 19 23 This problem is very simple since it is solved just by additions and subtractions. The solution is A B C D 1 7 2 10 4 23 2 9 1 3 11 24 3 5 12 6 8 31 21 15 19 23 These kind of Survo Puzzles are suitable as exercises when teaching basic arithmetics.

A B C D 1 * 2 6 * 33 2 4 * * 1 26 3 * 5 * 13 42 4 * 14 3 * 35 38 33 34 31 This puzzle is almost as easy as the first one since by plain additions and subtractions we find A B C D 1 * 2 6 * 33 2 4 12 9 1 26 3 8 5 16 13 42 4 * 14 3 * 35 38 33 34 31 The missing four numbers are 7,10,11, and 15. In the first row the sum of missing numbers is 33-2-6=25 and this can be presented as a sum of missing numbers only as 25=10+15. In the first column (A) the sum of missing numbers is 38-4-8=26 and this can be presented as a sum of missing numbers only as 26=11+15. Therefore A1 must be 15 and thereafter it is easy to complete the solution A B C D 1 15 2 6 10 33 2 4 12 9 1 26 3 8 5 16 13 42 4 11 14 3 7 35 38 33 34 31

A B C D 1 * * * * 26 2 * * * * 33 3 * * * * 19 31 7 25 15 It is rather easy to solve this problem by direct reasoning. The best starting point is column B since the only way to present number 7 as a sum (partition) of three distinct integers is 7=1+2+4. By similar arguments it can be deduced stepwise that columns A,B,C,D can have only the following partitions: numbers in use: B: 7=1+2+4 1 2 4 D: 15=3+5+7 3 5 7 A: 31=12+11+8=12+10+9 Number 12 is thus in column A. C: 25=10+9+6 6 9 10 A: 31=12+11+8 8 11 12 A B C D 1 * * * * 26 2 * * * * 33 3 * * * * 19 31 7 25 15 The unique partitions of columns are in the increasing order of numbers: 8 1 6 3 11 2 9 5 12 4 10 7 gives 33 as the sum. This must then be row 2! Thereafter it is easy to solve rows 1 and 3: Number 8 cannot be in row 1 because 8+2+9+5=24<26 (too small). Then A1=11 and the other numbers in row 1 must be 1,9,5 in this order. The remaining numbers are then 8,2,6,3 and they form row 3. The solutions is A B C D 1 11 1 9 5 26 2 12 4 10 7 33 3 8 2 6 3 19 31 7 25 15

A B C D E F G 1 * * * * * * * 77 2 * * * * * * * 28 3 * * * * * * * 126 36 39 30 33 42 27 24 It is essential to notice that the smallest seven numbers have the sum 1+2+3+4+5+6+7=28 and then they must be located in row 2 in some order. Similarly, since this is a 3 x 7 Survo puzzle, the sum of the seven largest numbers is 21+20+19+18+17+16+15=126 and they form the row 3 in some order. Then the remaining numbers 8,9,10,11,12,13,14 are in the row 1 in some order. The sum of the smallest numbers in these three sets is 1+8+15=24. This is the sum of column G and it must then contain these numbers in order 8,1,15. When noticing that sums of columns form an arithmetic sequence 24,27,30,33,36,39,42 it becomes clear how the other numbers of these three sets must be located on columns and the solution is A B C D E F G 1 12 13 10 11 14 9 8 77 2 5 6 3 4 7 2 1 28 3 19 20 17 18 21 16 15 126 36 39 30 33 42 27 24

Open 4 x 4 puzzles can be really hard and often the easiest way is to solve them by the swapping method. However, when using this techique one cannot be assured that the solution found is unique. A B C D 1 * * * * 49 2 * * * * 23 3 * * * * 34 4 * * * * 30 40 52 10 34 Problem 5 happens to be rather easy since its solution proceeds through systematic, logical steps as follows. The only possible partition of column C is 10=1+2+3+4. Let's tudy the first row. The three largest numbers 14,15,16 must be there, since 49-14-15-16=4 and then C1=4. We have A B C D 1 14r 15r 4 16r 49 2 * * 1c * 23 3 * * 2c * 34 4 * * 3c * 30 40 52 10 34 where, for example, 14r means that number 14 is in this row but its column is unknown so far. Similarly 1c means that number 1 is in this column. The column B needs large numbers. 13,12,11 are the biggest available and since 52-13-12-11=16, B1 must be 16 and we have A B C D 1 14r 16 4 15r 49 2 * 11c 1c * 23 3 * 12c 2c * 34 4 * 13c 3c * 30 40 52 10 34 The row 2 needs small numbers. Lowest available are 5,6 and since 23-5-6=12 we must have B2=11 and C2=1: A B C D 1 14r 16 4 15r 49 2 5r 11 1 6r 23 3 * 12c 2c * 34 4 * 13c 3c * 30 40 52 10 34 The largest available numbers for column A are 9,10 and since 40-10-9=21, we must have A1=15 and A2=6 making rows 1 and 2 ready: A B C D 1 15 16 4 14 49 2 6 11 1 5 23 3 9c 12c 2c * 34 4 10c 13c 3c * 30 40 52 10 34 The smallest available number is 7 and the only possibility for getting sum 30 for line 4 is to use numbers 9+12+2+7=30. Then the final solution is A B C D 1 15 16 4 14 49 2 6 11 1 5 23 3 10 13 3 8 34 4 9 12 2 7 30 40 52 10 34

A B C D E F G H I J 1 * * * * * * * * * * 64 2 * * * * * * * * * * 146 10 11 14 17 20 22 23 27 31 35 This Survo puzzle appeared in June 2006 and several correct solutions were presented. The following solution was given by Olli Mustonen and it is maybe the best one. Let's study the puzzle in four parts: A B C D E F G | H I J A - G | H - J 1 * * * * * * * | * * * 64 1 X11 | X12 64 --------------------------------------- = ----------------------- 2 * * * * * * * | * * * 146 2 X21 | X22 146 10 11 14 17 20 22 23 | 27 31 35 10 - 23 | 27 - 35 X22 seems to hold largest numbers, X11 smallest ones, and X12 and X21 seem to share numbers of medium size. What is the smallest possible sum for the three numbers in X12? The answer is obtained by placing the three largest numbers 18,19,20 in X22 - the order of them is irrelevant in respect to the sum of numbers in X12 and this sum is then 27+31+35-18-19-20=36. When this smallest possible sum is subtracted from the total sum 64 of row 1, the largest possible sum 28 of seven numbers in X11 is obtained, but this is also the sum of the smallest possible numbers 1+2+3+4+5+6+7=28. Thus we know now that X11 consists of numbers 1-7 in some order and the sums of numbers in parts X12 and X21 are 36 and 146-(18+19+20)=89, respectively. H2 cannot be 20 since this would give H1=7, but number 7 already appears somewhere in X11. The possible solutions for X12 are H1 (8,9), I1 (11,12,13), and J1 (15,16,17). Since 10 and 14 do not appear among these numbers, they have to be located in X21. We notice that A2 and B2 must be small. Is it possible to manage without the smallest free number 8? The next available numbers are 9 and 10 - A2 could be 9 and B2 10, but then both A1 and B1 would be 1 which is impossible. The conclusion is that number 8 is needed in X21 and cannot be in X12. Then H1 must be 9 and H2=18. Then also A2=8 and A1=2. Similarly: B2=10 and B1=1, C1=3 and C2=11. Since now number 11 is excluded from possible values of I1, I2 cannot be 20. The only alternative is I2=19 and then I1=12. This implies that J2=20 and J1=15. Remaining numbers in D2,E2,F2, and G2 are 13,14,16, and 17. Since no smaller number than 4 is available, it has to be put in D1 and then D2=13. From remaining numbers only 14+6=20 - thus E1=6 and E2=14. The only solutions for remaining cells are F1=5, F2=17, G1=7, and G2=16 and the final solution is A B C D E F G H I J 1 2 1 3 4 6 5 7 9 12 15 64 2 8 10 11 13 14 17 16 18 19 20 146 10 11 14 17 20 22 23 27 31 35

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Updated 2009-07-28 by webmaster'at'survo.fi.

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