There is an old adage that one person’s “creativity” is another person’s “cheating”. This week’s puzzle will test which side of the fence you sit on.

The numbers 1 to 9 have been written on cards and left on a table: the left-hand column adds up to 21 and the right one to 24. Move just one card so that the two columns add up to the same total. There’s a classic “Aha!” solution to this puzzle, but my daughter came up with a solution I wasn’t expecting. Since then, I have been offered at least 10 more distinct solutions.

How many solutions can you find that you regard as “creative” rather than “cheating”?

Answer next week

#120 More Catch up

Yes, Ann can guarantee victory. A tower of 23 or more will win the game for Ann, so she needs to add at least 15 to her current tower of height 8.

If Bob takes stack 9 (call the move “B9”), Ann plays 7 (A7), and can follow that with A8 or A4 & A6 depending on Bob’s next move. Likewise, she can force a win with the following moves: B8-A9 ; B7-A6; B6-A8. The only other option for Bob is B4 then B6/7/8/9, which Ann can follow with A9 or (if 9 is already taken), A8. It is deceptively difficult to develop a general strategy for this game.

Our football league archivist has another challenge on his hands (see Puzzle #104, 13 March). A second old newspaper report has surfaced, this time from the 1991 season. The table shows how things stood at the end of the season after all teams had played each other once. There were three points for a win and one for a draw, as you might expect, with Albion winning the league and Rovers bottom. How annoying that so much of the table got smudged.

Can you use your forensic skills to figure out all the match scores for that season?

P | W | L | D | GF | GA |

Points

Albion * * * * 4 * 7

United * * * * 4 5 *

Town * * * * 1 2 *

Rovers * * * * 1 3 1

Key: P = Played, W = Won, L = Lost, D = Drawn, GF = Goals For, GA = Goals Against, and * denotes an entry that has been smudged beyond recognition.

Solution next week

#115 A random robot

Solution

There is a 1/4 chance Roman the robot will finish where he started. There are four possible first moves he can take, and since he can only move perpendicular to his previous move, there are two possible moves for each of the last three steps. So there are 4脳2脳2脳2 = 32 possibilities.

To arrive back at the start, he has four possible first moves, two possible second moves for each of the first, and then, in order to cancel out the effect of the first two moves, there is only one option for each of the last two moves. So there are 4脳2脳1脳1 = 8 possible ways to return to his start point. This means 8/32 or 1/4 is the chance of this happening if all moves are random.