Sudoku E1 Trivia Quiz

Answer: 5

Row 4 has 5 blank squares which must take the numbers 1, 4, 5, 6 and 9. Column 9 already has 1, 4, 6 and 9, meaning only 5 can go in the red square on row 4 at column 9.

The same reasoning applies to the 7 which goes on row 6 at column 9.

Answer: 7

The bottom three clusters comprise rows 7, 8 and 9. Looking at rows 7 and 8, there are 7s are already in place. This means that the 7 on row 9 must be in column 8.

Looking columns 4, 5 and 6, there are 7s in columns 5 and 6 so the remaining 7 must go on row 5 in column 4. The same logic applies to the placement of a 6 on row 4 at column 3. Likewise, the 8s in columns 4 and 5, coupled with the 8 in row 3, mean the 8 on row 2 goes in column 6. The same reasoning gives you 1s on row 2 at column 1 and row 8 at column 6. We can also now deduce that the 7 on row 2 goes on column 7.

Answer: 7

From the reasoning in Q2 we know that the 7 in column 7 is on row 2. This in turn means that the red square in column 3 has a 7 since there are already 7s in columns 1 and 2.

Alternatively, with the numbers filled in so far, the only number missing out of the range 1 - 9 from row 3 and column 3, and the 3 x 3 grid the red square sits in, is 7. Using the same technique, we can also place a 5 on row 2 at column 3. Another iteration and we can place 8 on row 1 at column 3.

Answer: 1

Looking at the row 4 intersections with the various columns, the only place a 1 can go is on the red square.

In the same 3x3 grid, the 2 must go on row 5 as there are already 2s on rows 4 and 6. The only column free is column 6 for this 2. The same reasoning applies to the 5 on row 7 at column 6, there being 5s on columns 4 and 5. Similarly, the only free square for the 2 in the top right grid is in column 9 as there is already a 2 in each of row 1 and column 8.

Answer: 5

Since we worked out at Q4 that there is a 5 on row 7 at column 6, the 5 in the bottom right 3x3 grid must be in row 8. With 5s in columns 1 and 3, the 5 in column 2 must therefore go on the red square on row 9.

Answer: 3

With the last few numbers placed, there is now only one space left on column 9, namely the red square. The missing number is 3.

Looking at column 6, only two numbers are missing - 4 and 9. Row 3 has a 9 so the intersection between them has a 4. Similarly, row 2 is missing the numbers 3 and 6. With 3 already appearing on column 4, the intersection must have a 6. Looking at row 6, the missing numbers are 1, 4, 8 and 9. Column 5 already has 1, 8 and 9, leaving 4 for the intersection.

Answer: 3

In the top centre 3x3 grid, only the 2 and 3 are missing. Row 1 already has a 2 so the 3 goes in the red square.

There is a 3 missing from row 2 which goes in column 8. There is also a 9 missing from column 6 which goes in row 4. On row 6, the numbers 1, 8 and 9 are missing. Column 7 supplies 1 and 8, meaning the intersection holds a 9.

Answer: 6

On row 3, the missing numbers are 2, 3 and 6. The 2 is the only number missing from the top centre 3x3 grid so it goes in column 5. There is already a 3 in the top right grid, so the 3 goes in column 2. That leaves the number 6 for the red square with a 3 on column 2 completing that line.

We can also put a 6 on row 1 at column 2 (last number in that 3x3 grid) and a 4 on row 4 at column 7 (the final number on row 4). Similar technique allows the placement of a 5 on row 1 at column 7, a 3 on row 5 at column 7, a 5 on row 8 at column 8, and a 6 at the bottom of column 5.

Answer: 8

The last few number placements leave a 9 to complete row 1 at column 8. Similarly, the missing number from row 9 and column 4 is a 4 which goes at the intersection. This leaves 6 and 8 missing from column 1, however there is already a 6 in the left-hand central 3x3 grid so the red square contains the 8.

We can also place a 9 at the intersection of row 9 with column 2, being the missing number.

Answer: 4

The existing 4s in rows 7 and 9 mean that the red square on row 8 can only have a 4. The remaining numbers can be filled in using the same techniques, leading to the correct solution which is:

2 6 8 - 1 3 7 - 5 9 4
1 4 5 - 6 9 8 - 7 3 2
9 3 7 - 5 2 4 - 8 6 1
- - - - - - - - - - - - -
3 7 6 - 8 1 9 - 4 2 5
8 9 4 - 7 5 2 - 3 1 6
5 1 2 - 3 4 6 - 9 8 7
- - - - - - - - - - - - -
7 2 1 - 9 8 5 - 6 4 3
6 8 3 - 4 7 1 - 2 5 9
4 5 9 - 2 6 3 - 1 7 8