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It is noted that any two members have at least one favorite movie in common.

What is the largest number of members that have necessarily selected a same movie?

Prove that.

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- Thread starter Rogerio
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- #1

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It is noted that any two members have at least one favorite movie in common.

What is the largest number of members that have necessarily selected a same movie?

Prove that.

- #2

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mathal

- #3

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Possibly because it wasn't worded well. Poorly worded puzzles tend to be ignored.I'm not sure why this wasn't solved sooner.

Here's my interpretation:

Each members of a 23-member moviegoers' club selects two movies from a list of 50 movies as their personal favorite movies. On examining the selections, it is noted that the selections by any two members have at least one movie in common. Given this, what is the smallest possible number of members who voted for the most popular movie, the movie that was selected by the most number of members?

Note: There may be a tie for the most popular movie.

Note: There may be a tie for the most popular movie.

TL;DR answer:

Sixteen

Details:

The set U comprises at least two members (all members voted for the same two movies) and at most 24 members (all members voted for the same most popular movie, but their is no commonality amongst their second choices). In both of these extreme cases, the number of people who voted for the most popular movie is 23. In fact, if the cardinality of U is four or more then all of the members had to agree on one movie as the most popular. Thus P is 23 if the cardinality of U is anything but three. The only way to have P less than 23 is if the cardinality of U is three.

Designate these three movies as A, B, and C, and the designate the number of people who voted for movies A & B as

- #4

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A member has two favorite movies. One or both match one or both of the other member's choices. In the extreme case the other 22 members are split equally between the two movies the member chose. 12 is the answer. (11+1) Any higher number is not necessarily what happened. 12 always will be true.

mathal

Hi Mathal,

there is at least a movie that received more than 12 votes, necessarily.

So, 12 is not the answer.

- #5

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Possibly because it wasn't worded well.Poorly worded puzzles tend to be ignored.

Here's my interpretation:

Each members of a 23-member moviegoers' club selects two movies from a list of 50 movies as their personal favorite movies. On examining the selections, it is noted that the selections by any two members have at least one movie in common. Given this,what is the smallest possible number of members who voted for the most popular movie, the movie that was selected by the most number of members?

That's right, the way puzzles are worded matters a lot!

And it seems you have correctly understood my puzzle.

However, the answer to your version (which has nothing to do with the original idea) is "ONE" .

If, for instance, 23 members had voted for a movie, then there would be one member who had voted for that movie.

And 1 is smaller then 23.

So, "one" is the smallest number of members who voted for the most popular movie.

Come on, this is not a proof.In fact, if the cardinality of U is four or more then all of the members had to agree on one movie as the most popular. Thus P is 23 if the cardinality of U is anything but three. The only way to have P less than 23 is if the cardinality of U is three.

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- #6

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Nonsense. You already said that that movie received 23 votes.However, the answer to your version (which has nothing to do with the original idea) is "ONE" .

If, for instance, 23 members had voted for a movie, then there would be one member who had voted for that movie.

And 1 is smaller then 23.

So, "one" is the smallest number of members who voted for the most popular movie.

This apparently is a minimax problem. You worded it very poorly. I had to read it multiple times before I realized that you meant smallest rather than largest.

Sure it is. Read the proofs in an advanced math text. They skip over the obvious steps. If four (or more) movies received a vote by at least one member, then there is exactly one movie in that set of four (or more) movies for which each of the 23 members voted. That's rather obvious, isn't it?Come on, this is not a proof.

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And this is not "advanced math" - it should be found in an introductory logic text.

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In fact, if the cardinality of U is four or more then all of the members had to agree on one movie as the most popular. Thus P is 23 if the cardinality of U is anything but three. The only way to have P less than 23 is if the cardinality of U is three.

If four (or more) movies received a vote by at least one member, then there is exactly one movie in that set of four (or more) movies for which each of the 23 members voted. That's rather obvious, isn't it?

Saying it is obvius is not a proof.

Prove it.

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I'm not sure why this wasn't solved sooner.

Hi Mathal,

I don't know, either.

But I would like to point out that "the largest necessary" is equivalent to "the smallest enough".

Maybe someone doesn't understand this.

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- #10

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Some advice: When someone tells you that what you wrote isn't clear, you most likely did not write clearly. Don't be defensive. Peer review is the time to "check your ego at the door." If you (or anyone else) have a better wording than mine, have at it.

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mfb

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That is easy to prove.Saying it is obvius is not a proof.

Prove it.

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