Nine square tiles are laid in a 3 × 3 grid. The cells are numbered 1 to 9 row by row, as shown:
| | Column 1 | Column 2 | Column 3 |
|---|---|---|---|
| Row 1 | 1 | 2 | 3 |
| Row 2 | 4 | 5 | 6 |
| Row 3 | 7 | 8 | 9 |
Each tile is painted in exactly one of four colours: Red, Green, Blue or Yellow. The colouring satisfies the following conditions:
1. No two tiles that share an edge have the same colour. (Tiles that touch only at a corner may have the same colour.)
2. Exactly three tiles are Red, and exactly two tiles each are Green, Blue and Yellow.
3. Every Red tile is in a corner cell.
4. Both Blue tiles are in the same row.
5. The tile in cell 9 is Yellow.
| | Column 1 | Column 2 | Column 3 |
|---|---|---|---|
| Row 1 | 1 | 2 | 3 |
| Row 2 | 4 | 5 | 6 |
| Row 3 | 7 | 8 | 9 |
Each tile is painted in exactly one of four colours: Red, Green, Blue or Yellow. The colouring satisfies the following conditions:
1. No two tiles that share an edge have the same colour. (Tiles that touch only at a corner may have the same colour.)
2. Exactly three tiles are Red, and exactly two tiles each are Green, Blue and Yellow.
3. Every Red tile is in a corner cell.
4. Both Blue tiles are in the same row.
5. The tile in cell 9 is Yellow.
Question 1
Single correct option
What is the colour of the tile in cell 5?
- A. Blue
- B. Green
- C. Yellow
- D. Red
Show answer and solution
Answer: (C) Yellow
Deduction: Reds occupy three of the four corners (condition 3); cell 9 is Yellow, so Reds are in cells 1, 3 and 7.
Row 1 has only cell 2 free and Row 3 has only cell 8 free, so the two Blues (same row) must be in Row 2. Cells 4, 5 are adjacent and so are 5, 6, so the Blues are in cells 4 and 6.
Cells 2, 5 and 8 receive Green, Green and Yellow. Cell 5 touches both 2 and 8, so it cannot share a colour with them: cell 5 = Yellow, cells 2 and 8 = Green.
| | C1 | C2 | C3 |
|---|---|---|---|
| R1 | Red | Green | Red |
| R2 | Blue | Yellow | Blue |
| R3 | Red | Green | Yellow |
Row 1 has only cell 2 free and Row 3 has only cell 8 free, so the two Blues (same row) must be in Row 2. Cells 4, 5 are adjacent and so are 5, 6, so the Blues are in cells 4 and 6.
Cells 2, 5 and 8 receive Green, Green and Yellow. Cell 5 touches both 2 and 8, so it cannot share a colour with them: cell 5 = Yellow, cells 2 and 8 = Green.
| | C1 | C2 | C3 |
|---|---|---|---|
| R1 | Red | Green | Red |
| R2 | Blue | Yellow | Blue |
| R3 | Red | Green | Yellow |
Question 2
Single correct option
How many tiles share an edge with at least one Blue tile?
- A. 5
- B. 4
- C. 3
- D. 6
Show answer and solution
Answer: (A) 5
Deduction: Reds occupy three of the four corners (condition 3); cell 9 is Yellow, so Reds are in cells 1, 3 and 7.
Row 1 has only cell 2 free and Row 3 has only cell 8 free, so the two Blues (same row) must be in Row 2. Cells 4, 5 are adjacent and so are 5, 6, so the Blues are in cells 4 and 6.
Cells 2, 5 and 8 receive Green, Green and Yellow. Cell 5 touches both 2 and 8, so it cannot share a colour with them: cell 5 = Yellow, cells 2 and 8 = Green.
| | C1 | C2 | C3 |
|---|---|---|---|
| R1 | Red | Green | Red |
| R2 | Blue | Yellow | Blue |
| R3 | Red | Green | Yellow |
Blue tiles are in cells 4 and 6. Cell 4 touches 1, 5, 7; cell 6 touches 3, 5, 9. Union = {1, 3, 5, 7, 9} → 5 tiles.
Row 1 has only cell 2 free and Row 3 has only cell 8 free, so the two Blues (same row) must be in Row 2. Cells 4, 5 are adjacent and so are 5, 6, so the Blues are in cells 4 and 6.
Cells 2, 5 and 8 receive Green, Green and Yellow. Cell 5 touches both 2 and 8, so it cannot share a colour with them: cell 5 = Yellow, cells 2 and 8 = Green.
| | C1 | C2 | C3 |
|---|---|---|---|
| R1 | Red | Green | Red |
| R2 | Blue | Yellow | Blue |
| R3 | Red | Green | Yellow |
Blue tiles are in cells 4 and 6. Cell 4 touches 1, 5, 7; cell 6 touches 3, 5, 9. Union = {1, 3, 5, 7, 9} → 5 tiles.
Question 3
Numerical answer
If condition 4 (both Blue tiles in the same row) were removed, all other conditions remaining, how many different colourings of the grid would be possible?
Show answer and solution
Answer: 6
Reds are still forced into cells 1, 3, 7 and cell 9 is Yellow. The remaining cells 2, 4, 5, 6, 8 get Green, Green, Blue, Blue, Yellow.
Cell 5 touches all of 2, 4, 6 and 8. If cell 5 were Green (or Blue), the other Green (Blue) tile would have to be in one of 2, 4, 6, 8 and would touch cell 5 – not allowed. So cell 5 = Yellow.
Cells 2, 4, 6, 8 touch each other only through corners, and each touches only Reds/Yellow otherwise, so Green, Green, Blue, Blue can be placed in them in any way: C(4, 2) = 6 colourings.
Cell 5 touches all of 2, 4, 6 and 8. If cell 5 were Green (or Blue), the other Green (Blue) tile would have to be in one of 2, 4, 6, 8 and would touch cell 5 – not allowed. So cell 5 = Yellow.
Cells 2, 4, 6, 8 touch each other only through corners, and each touches only Reds/Yellow otherwise, so Green, Green, Blue, Blue can be placed in them in any way: C(4, 2) = 6 colourings.
Question 4
Numerical answer
If condition 5 (cell 9 is Yellow) were removed, all other conditions remaining, how many different colourings of the grid would be possible?
Show answer and solution
Answer: 8
Reds occupy any three of the four corners – 4 choices; call the free corner X.
The two Blues must share a row with a gap between them. In the outer row containing X, the free cells are X and the middle-edge cell next to it – adjacent, so impossible; the other outer row has only one free cell. Hence the Blues are always in cells 4 and 6.
The remaining cells 2, 5, 8 and X receive Green, Green, Yellow, Yellow, and equal colours must not touch. Among these four cells the adjacencies form a chain: X touches one of 2/8, which touches 5, which touches the other of 2/8 (e.g. for X = 9: 9–8–5–2). The only way to split a 4-cell chain into two non-touching pairs is {X, 5} and {2, 8}. Either pair can be Green (the other Yellow) → 2 colourings.
Total = 4 corners × 2 = 8. (Condition 5 is exactly what fixes X = 9 and makes cell 5 Yellow.)
The two Blues must share a row with a gap between them. In the outer row containing X, the free cells are X and the middle-edge cell next to it – adjacent, so impossible; the other outer row has only one free cell. Hence the Blues are always in cells 4 and 6.
The remaining cells 2, 5, 8 and X receive Green, Green, Yellow, Yellow, and equal colours must not touch. Among these four cells the adjacencies form a chain: X touches one of 2/8, which touches 5, which touches the other of 2/8 (e.g. for X = 9: 9–8–5–2). The only way to split a 4-cell chain into two non-touching pairs is {X, 5} and {2, 8}. Either pair can be Green (the other Yellow) → 2 colourings.
Total = 4 corners × 2 = 8. (Condition 5 is exactly what fixes X = 9 and makes cell 5 Yellow.)
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