This week’s puzzle is based on a game that was originally described in the "Mathematical Games" section of Scientific American, April, 1969 by Anatol W Holt.

Two players each think of a Target Word containing the agreed number of letters. It’s best to start with three-letter words, and increase the length as the players improve. Each player in turn suggests a Probe Word, and receives a reply of Odd or Even. This response indicates the number of correct letters in the correct place in the Probe Word, with zero being even. The first player to find their opponent’s Target Word is the winner.

This is quite a fun game to play in its own right, and it bears a certain resemblance to a board game called Mastermind, sold by Hasbro. This in turn also resembles an old pen-and-paper game called Bulls and Cows.

All three of these games have been implemented on various computers, phones and tablets, and I’m sure that some of our readers will be thinking of writing a program for one of the games.

The puzzle is to determine the Target Word from the six Probe Words and Responses below. It can be solved by trial and error (or by writing a program), but the challenge here is to deduce the word logically. From the information given you can work through the problem and find the solution.

DAY – Even

MAY – Even

BUY – Even

SAY – Odd

DUE – Odd

TEN – Odd

As usual you can post the answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answers at 9.00 on Monday

Liz JonesWhat Rich said :)

Rich TwoseSo…

If we take the first two terms, they are both even, and so either have 0 or 2 correct letters. If they have 2 correct letters, they must be AY. The third term though also has either 0 or 2 correct, and doesn’t feature the A from AY. This means that if it contains 2 correct terms, it must be the B-Y. Combine these, and you get BAY.

When you reference this against term 4 though (with an odd value, and therefore containing either one or three correct values), it conflicts. This proves that the even value for terms 1, 2 and 3 must be because they contain 0 correct terms, not 2.

From here, it is fairly straight forward. Term 4 has either 1 or 3 correct letters, but we know the AY is wrong, so it must be 1 character (the S). Term 5 is the same, but we know the DU doesn’t fit, so the final E must be correct. On the final word, all three letters are viable, but only the middle character needs to be filled, and so it is E. This gives us SEE.