15 Logic
It was a standard procedure, particularly in the early days, for us to put in the AMC each year what we would call a logic problem. The most famous, and quite typical, was the following, from the 1984 paper.
Example 15.1
Albert, Bernard, Charles, Daniel and Ellie play a game in which each is a frog or kangaroo. Frogs' statements are always false while kangaroos' statements are always true.
Albert says that Bernard is a kangaroo.
Charles says that Daniel is a frog.
Ellie says that Albert is not a frog.
Bernard says that Charles is not a kangaroo.
Daniel says that Ellie and Albert are different kinds of animals.
How many frogs are there?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Solution 15.1
Let a single-headed arrow represent `... says ... is a kangaroo' and a double-headed arrow represent `... says ... is a frog'. Then we have
![[Diagram]](https://lh5.googleusercontent.com/VZy20txZzSRgt5TTLs8QiK7xtS8TGQz6939lCxqTXsRcsFWZks07LU5SUdo3FXYJebA5n_RloQulkfEgO3o_TxvdE7jf2V_nzFVXi9T0DwN8JWkG=w1280)
Assume Ellie is a kangaroo, and hence that his statement is true. Thus
Albert is a kangaroo,
Bernard is a kangaroo,
Charles is a frog and
Daniel is a kangaroo.
But this is not possible since Ellie and Albert are then both kangaroos, contrary to Daniel's statement. This proves that Ellie is not a kangaroo, but a frog instead. Being a frog, Ellie's statement is false. Thus
Albert is a frog,
Bernard is a frog,
Charles is a kangaroo and
Daniel is a frog.
There are 4 frogs, hence the solution is (D).
Comment
![[Diagram]](https://lh4.googleusercontent.com/LxvZk_Iw8K0NCk9Blu76ZtgYnhdJIYd0K2Ghw_yHp2PDhcmk7E5rX2swEmtKnwUsihl2LrhO9Hi1WWCRM5ceLO1VfLJRgRjeMyQvA_VGWHNX50ll=w1280)
We had some fun with our French colleagues and combined with them to arrange for the kangaroos to tell lies and frogs to tell the truth in the French translation used in New Caledonia and French Polynesia.
More advanced logic problems
Logic can lie behind puzzles which circulate the wider community. My attention in 2015 was drawn by AMT Board Chair Greg Taylor to the following problem which was posted on a web site for the Singapore and Asian Schools Mathematical Olympiad. By a curious coincidence, the two characters whose names start with A and B are the same in both the problem above and the one below.
Example 15.2
Albert and Bernard became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.
15 May, 16 May, 19 May.
17 June, 18 June.
14 July, 16 July.
14 August, 15 August, 17 August.
Cheryl then confides with Albert the month of her birthday and with Bernard the day of her birthday.
Albert says `I don’t know when Cheryl’s birthday is, but I know that Bernard does not know either'.
Bernard replies `At first I didn’t know when Cheryl’s birthday is, but I know now.
Albert replies `In that case I also know when Cheryl’s birthday is'.
So when is Cheryl’s birthday?
Solution 15.2
If Bernard had the number 18 or 19, which appear only once, he would know. For Albert to know that Bernard doesn’t have either of these numbers, Albert must have been given the month of either July or August.
So Bernard has one of the numbers corresponding to these months, namely 14, 15, 16 or 17. If he had 14 he wouldn't know if it was July or August. So he must have been given 15, 16 or 17, either of which would have been given the answer.
For Albert to know the answer now he must have been given July and the answer is 16 July. If he had been given August the date could still have been one of 15 or 17.