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Four Times Nine Equals Six Squared (Posted on 2006-10-16) Difficulty: 3 of 5
Each of the numbers 1 to 9 appears four times in the 6x6 grid. If a number appears twice in any row or column, it is mentioned in the relevant clue.
                 1  2  3  4  5  6
               +--+--+--+--+--+--+
            1  |  |  |  |  |  |  |
               +--+--+--+--+--+--+
            2  |  |  |  |  |  |  |
               +--+--+--+--+--+--+
            3  |  |  |  |  |  |  |
               +--+--+--+--+--+--+
            4  |  |  |  |  |  |  |
               +--+--+--+--+--+--+
            5  |  |  |  |  |  |  |
               +--+--+--+--+--+--+
            6  |  |  |  |  |  |  |
               +--+--+--+--+--+--+
ACROSS
1. Total 22.
2. The numbers in the first three cells are even; those in the last three are odd.
3. Two 5s, one of which is enclosed by two 1s. Total 27.
4. 7 and 9 are the only odd numbers.
5. No 2s, 3s or 9s.
6. Two 3s enclosing an 8. No 7s. Total 31.

DOWN
1. No 2s, 6s or 7s.
2. Two 7s separated by two squares. Four even numbers.
3. The sum of the lowest three cells is 14.
4. Total 21.
5. No 3s, 4s or 7s.
6. Two 9s separated by two squares. Total 39.

See The Solution Submitted by Josie Faulkner    
Rating: 4.3846 (13 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution - with very long explaination!! | Comment 11 of 22 |

I had a bit of confusion with clue 3 Down with 'lowest' and if it meant the three cells with the lowest valued numbers in or if it meant the bottom three cells. Eventually worked out it meant the bottom three cells as other way didnít work!!

Sorry for the long explaination!!

1) In column 2 we know row 2 and row 6 do not have any 7s (clues 2A and 6A). Therefore the two 7s are in rows 1, 3, 4 or 5. We know there are two squares between the 7s (clue 2D) so they must be in rows 1 and 4.

- 7 - - - -

- - - - - -

- - - - - -

- 7 - - - -

- - - - - -

- - - - - -

2) Row 1 adds up to 22 (clue 1A) so the numbers in row 1 is 1, 2, 3, 4, 5 and 7.

3) Column 6 has two 9s but from 2) above and clue 5A the 9s are in either rows 2, 3, 4 or 6. We know there are two squares between the 9s (clue 2D) so they must be in rows 3 and 6.

- 7 - - - -

- - - - - -

- - - - - 9

- 7 - - - -

- - - - - -

- - - - - 9

4) Row 3 has two 1s enclosing a 5. We know column 6 is a 9 and column 2 is even (clue 2A) so the two 1s must be in columns 3 and 5 with the 5 in column 4. The other 5 canít be in column 2 so it must be in column 1. We also know the sum of the row is 27 so the number in column 2 must be 6.

- 7 - - - -

- - - - - -

5 6 1 5 1 9

- 7 - - - -

- - - - - -

- - - - - 9

5) Row 6 has two 3s enclosing an 8 (clue 6A). We know there is no 3 in column 2 as has to be even (clue 2D) or column 5 (clue 3D) and column 6 is a 9 so the two 3s must be in columns 1 and 3 with 8 in column 2.

- 7 - - - -

- - - - - -

5 6 1 5 1 9

- 7 - - - -

- - - - - -

3 8 3 - - 9

6) Column 2 row 5 cell is even i.e. 2, 4, 6 or 8 (clue 2D), it canít be a 2 (clue 5A) and canít be a 6 or 8 as there is already a 6 and 8 in column 2 so it must be a 4. The remining cell in that row must be the remaining even number, 2.

- 7 - - - -

- 2 - - - -

5 6 1 5 1 9

- 7 - - - -

- 4 - - - -

3 8 3 - - 9

7) we know in column 1 the other 4 numbers must be 1, 4, 8 and 9 (clue 1D). Number 9 can't be in row 1 (from 2. above), row 2 (clue 2A), or row 5 (clue 5A) so it must be in row 4. The other 4 numbers in this row must be even (clue 4A).

- 7 - - - -

- 2 - - - -

5 6 1 5 1 9

9 7 - - - -

- 4 - - - -

3 8 3 - - 9

8) The cell in column 5 and row 2 is odd but it can't be 5 as the four 5s are in rows 1 (from 2. above), two in row 3, and 1 in row 5 (clue 2A). It can't be 1 as 1 is already in that column and it canít be 3 or 7 (clue 5D), therefore it must be 9.

- 7 - - - -

- 2 - - 9 -

5 6 1 5 1 9

9 7 - - - -

- 4 - - - -

3 8 3 - - 9

9) We know that the sum of column 4 is 21 so this column must have the numbers 1-6. We know the number in row 2 is odd (clue 2A) so must be either 1, 3 or 5. It can't be a 5 as already a 5 in that column. We know it canít be a 1 as the four ones are in row 1 (clue 1A), two in row 3 and in row 5 (clue 5A), therefore it must be a 3.

- 7 - - - -

- 2 - 3 9 -

5 6 1 5 1 9

9 7 - - - -

- 4 - - - -

3 8 3 - - 9

10) We know the cell in column 6 row 2 is odd (clue 2A) but it canít be 3 or 9 as already in its row. It also canít be a 1 or 5 as these are one of each in rows 1 and 5, and 2 of each in row 3, therefore it must be a 7.

- 7 - - - -

- 2 - 3 9 7

5 6 1 5 1 9

9 7 - - - -

- 4 - - - -

3 8 3 - - 9

11) We know there must be a 7 in row 5 (clue 5A), it can't be in column 6 as one is already there. It can't be in columns 1 or 5 (clue 1D and 5D), and it can't be in column 4 as they only have the numbers 1-6, therefore it must be in column 3. We also know that the sum of the lowest three cells is 14 so the cell above this one must be 4.

- 7 - - - -

- 2 - 3 9 7

5 6 1 5 1 9

9 7 4 - - -

- 4 7 - - -

3 8 3 - - 9

12) We know the last cell in column 4 must be a 1, 2, 4 or 6 and the last cell in column 5 is a 2, 5 or 6. The sum of row 6 is 31 so the two cells in columns 4 and 6 must add up to 8. The only way this can be done is by either of them being a 2 or a 6.

13) We know they are 4s in rows 1, 4 and 5. Rows 3 and 6 don't have any 4s so the forth 4 must be in row 2. Column 3 already has a 4 so the 4 must be in column 1.

- 7 - - - -

4 2 - 3 9 7

5 6 1 5 1 9

9 7 4 - - -

- 4 7 - - -

3 8 3 - - 9

14) The remaining two cells in column 1 must be 1 and 8. We know 8 isnít in row 1 so row 1 must have the 1 and row 5 the 8.

1 7 - - - -

4 2 - 3 9 7

5 6 1 5 1 9

9 7 4 - - -

8 4 7 - - -

3 8 3 - - 9

15) We know column 4 has a 1 in it. This canít be in row 1 as already has a 1, row 4 does not have any more odd numbers and row 6 is either a 2 or 6 so the 1 must be in row 5. Similarly there is a 4 in this column too. It canít be in row 6 as this is a 2 or 6 and row 4 already has a 4 so it must be in row 1.

1 7 - 4 - -

4 2 - 3 9 7

5 6 1 5 1 9

9 7 4 - - -

8 4 7 1 - -

3 8 3 - - 9

16) Row 1 contains a 3. This can't be in column 5 (clue 5D) or column 3 as it already has a 3 so it must be in column 6.

1 7 - 4 - 3

4 2 - 3 9 7

5 6 1 5 1 9

9 7 4 - - -

8 4 7 1 - -

3 8 3 - - 9

17) The cell in column 3, row 2 is either a 6 or 8 and we know the 4 6s are in the last 4 rows so row 2 can't be a 6 so must be 8.

1 7 - 4 - 3

4 2 8 3 9 7

5 6 1 5 1 9

9 7 4 - - -

8 4 7 1 - -

3 8 3 - - 9

18) We know the sum of the last column is 39 so the remaining two cells in this column must add up to 11. The cell in row 4 is either a 2, 6 or 8 and the cell in row 5 is 5 or 6. The only way these add to 11 is by row 4 being 6 and row 5 being 5. This also means the remining cell in row 5 must be 6.

1 7 - 4 - 3

4 2 8 3 9 7

5 6 1 5 1 9

9 7 4 - - 6

8 4 7 1 6 5

3 8 3 - - 9

19) The remaining cells in row 4 must be 2 or 8. 8 canít be in column 4 so 8 must be in column 5 and 2 in column 4. The remaing cell in column 4 is therefore 6.

1 7 - 4 - 3

4 2 8 3 9 7

5 6 1 5 1 9

9 7 4 2 8 6

8 4 7 1 6 5

3 8 3 6 - 9

20) The remaining cell in row 6 must be a 2. The remaining cells in row 1 are 2 or 5, 2 can't be in column 5 as already in this column so the 2 must be in column 3 and 5 in column 5.

1 7 2 4 5 3

4 2 8 3 9 7

5 6 1 5 1 9

9 7 4 2 8 6

8 4 7 1 6 5

3 8 3 6 2 9

 


  Posted by Lisa on 2006-10-19 07:47:46
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