Did you solve it? Intrigue at the Pet Hotel | Math

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Earlier today, I offered you these three puzzles from Japanese setter Tadao Kitazawa,

1. The Pet Hotel

At the Pet Hotel, the rooms are numbered from 1 to 5, in that order. Each room can accommodate an animal and has its own light. At night, a nervous animal leaves the light on. An animal that is not nervous turns off the light. Each of the rooms 1 to 5 is always occupied by a dog or a cat, and everyone leaves after one night.

a) On Saturday night, a dog is nervous if and only if there are cats in the two adjacent rooms. A cat is nervous if and only if there is a dog in at least one adjacent room. We observe that four rooms remain lit. How many cats are there at the Pet Hotel?

b) On Sunday evening, a dog is nervous if and only if there are other dogs in the two adjacent rooms. A cat is nervous if and only if there is another cat in at least one adjacent room. It is observed that only one room remains lit. How many cats are there at the Pet Hotel?

The solution a) 3 cats b) 2 cats

a) Consider three cases. Case 1, only room 3 is dark. Suppose there is a cat in this room. So rooms 2 and 4 must both have cats. Both of these cats are nervous, so 1 and 5 have dogs. None of these dogs are nervous, so 1 and 5 would be dark, which contradicts the premise of case 1. Suppose room 3 has a dog. So at least one of 2 and 4 has a dog. This dog cannot be nervous, so he would turn off the light, which again would lead to the contradiction.

Case 2, room 2 or 4 is dark. Let’s say it’s 2. Suppose there’s a cat in this room. So 1 and 3 must have cats. But if so, then the cat in 1 is not nervous, which leads to the contradiction. Suppose there is a dog in 2. Then 1 or 3 has a dog, but this dog will not be nervous, which will lead to a contradiction. If we started with 4, the same logic applies.

Case 3, room 1 or 5 is dark. Let’s say it’s 1. Suppose there’s a cat in this room. Then there is a cat in 2, which means there is a dog in 3 and a cat in 4. Room 5 can have a cat or a dog, but in either case it is not nervous and turn off the light. So suppose there is a dog in 1. If room 2 has a dog, it won’t be nervous, so the light goes out and we get a contradiction. Bedroom 2 therefore has a cat. If room 3 has a dog, then room 4 must have a cat, and anything in room 5 will turn off the light. Contradiction.

So, suppose 1 has a dog. Room 2 must be a cat as a dog in this room would not be nervous. If room 3 has a dog, it’s only nervous if room 4 has a cat, which will be nervous, but that means the one in room 5 isn’t nervous, leading to a contradiction. Thus room 3 has a cat, room 4 a dog and room 5 a cat. It works, and this is our solution – 3 cats.

b) Using a process similar to the one above, you will find a solution when the lit room is 3 (resulting in a cat/dog/dog/dog/cat solution). So 2 cats.

2. Shaken, not bumped

Among six children, every handshake is between a boy and a girl. Each of the four children shakes hands with exactly two others. Each of the other two shakes hands with exactly three others. Are these two children shaking hands?

The solution: Yes. First, determine how many boys and how many girls are in the group. There can only be one boy, because that would mean that only that boy can shake hands with more than one person (since handshakes are between boys and girls), and we know that everyone shakes the hand of more than one person. If there are exactly two boys, then the boys would be the children shaking hands with three others, and the four girls would each be shaking hands with two others. But that would mean each girl shaking hands with two boys, which means the boys each shaking hands with four people, which contradicts the question. So there are at least three boys. Repeating the above argument for girls, we deduce that there are at least three girls. We conclude that the group has three boys and three girls.

Now let’s see if the two children shaking hands with exactly three other people are of the same sex. Case 1 Let’s say they are, and let’s say they’re girls. Then everyone would shake hands with each boy, and each boy would already have his two assigned handshakes. The third girl won’t have a handshake, so it doesn’t work.

Case 2 The two children are of the opposite sex. If they are, they should shake hands. Here is a way to make it work If A, B, C are girls and X, YZ are boys, then there are handshakes between AX, AY, AZ, BX, CX, BY, CZ.

3. I should be so lucky

Three girls, Akari, Sakura, and Yui, are each given a positive integer, which they keep secret from each other. They are told that the sum of the numbers is 12. A girl is considered “lucky” if she has the highest number. It is possible that one, two or all three girls are “lucky”.

Akari says, “I don’t know who’s lucky.”

Sakura says, “I still don’t know who’s lucky.”

Yui says, “I still don’t know who’s lucky.”

Akari says, “Now I know who’s lucky!”

Who is lucky ?

The solution: Sakura and Yui are lucky.

If Akari does not know that she is lucky, we can deduce that her number is at most 5. This is because if she had 6, she would know that only she is lucky, since he would be impossible for others to have 6 or more.

Likewise, we can deduce that Sakura and Yui also have at most 5. After everyone speaks once, the three girls know that none of them have a number higher than 5.

They know that all numbers add up to 12. There are ten possible combinations of numbers 5 or less that add up to 12:

  • ASY

  • 5 5 2

  • 5 2 5

  • 2 5 5

  • 5 4 3

  • 5 3 4

  • 4 5 3

  • 4 3 5

  • 3 5 4

  • 3 4 5

  • 4 4 4

We can rule out the first case, because if so, Yui would have known the other two were lucky. There are three other cases where Akari has 5, three when she has 4, two when she has 3, and one when she has 2. Since she is able to deduce who is lucky, she must have 2. (Since if she had another number, she wouldn’t know exactly who was lucky). So, she has 2, the others have 5, and Sakura and Yui are lucky.

I hope you enjoyed these puzzles. I will be back in two weeks.

Thanks to Tadao Kitazawa for today’s puzzles. His book Arithmetical, Geometrical and Combinatorial Puzzles from Japan is full of puzzles like the ones above.

I install a puzzle here every two weeks on a Monday. I’m always on the lookout for great puzzles. If you want to suggest one, write to me.

I’ve authored several puzzle books, including the most recent Language Lover’s Puzzle Book. I also give school lectures on math and puzzles (online and in person). If your school is interested, please contact us.

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