PK 2018 Q24
24. Each of the numbers 1, 2, 3, 4, 5, 6 is to be placed in the cells of a 2 × 3 table, with one number in each cell. In how many ways can this be done so that in each row and in each column the sum of the numbers is divisible by 3?
- A 36
- B 42
- C 45
- D 48
- E another number
Pink Kangaroo 2018, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 48
Each column will contain two numbers that add to a multiple of 3. The only possible pair for 3 is 6 so one of these will appear in the top row, and the other on the bottom row. The numbers in the same row as 3 must also add to a multiple of 3, so could be 1, 2 or 1, 5 or 2, 4 or 4, 5. Once this row is decided, the other row is entirely determined by this choice; e.g. if the top row is 3, 1, 2, then the bottom row must be 6, 5, 4. Hence, ignoring order in the first instance, there are just these four choices for what goes in the rows.
| 3 | 1 | 2 |
| 6 | 5 | 4 |
| 3 | 1 | 5 |
| 6 | 2 | 4 |
| 3 | 2 | 4 |
| 6 | 1 | 5 |
| 3 | 4 | 5 |
| 6 | 2 | 1 |
For each of these, the rows can be swapped, which doubles the number of options to 8.
For each of these, the three columns can be arranged in 6 ways (3 choices for first column, 2 choices for second, so 3 × 2 = 6 choices). Hence there are 8 × 6 = 48 choices.
Solution © UK Mathematics Trust, from the PK 2018 solutions.