83 beach goers

Logic Level 3

$83$ people are at the beach, each holding a coin in one hand. $25$ have one in their right hand, and $58$ have one in their left hand, but you don't know which is which.

Your task is to split the people into two groups such that the same number in each group has a coin in their right hand, and you can do this in two steps:

Separate the people into two groups with $n$ people in the smaller group.
Tell $m$ people, including however many you like in either or both groups, to switch their coin to the opposite hand.

What is $m + n$ ?

Clarification: You have no idea which $25$ of the $83$ people has the stone in their right hand. You split the people into two groups of size $n$ and $83-n$ , (with no control over how many right handed or left handed coin holders are in each group) and then you ask $x \leq n$ people in the first group to switch (not knowing which hand holds the coin for any of them), and $y \leq 83-n$ to switch in the other group, again not knowing which of the people you ask might have a coin in their left or right hand.

If you think this is impossible to do, provide -1 as your answer.

The answer is 50.

1 solution

Geoff Pilling
Mar 9, 2017

The solution is to separate them into groups of size $58$ and $25$ and tell all the people in the group of $25$ to switch the coin to their other hand.

After separating them, lets say there are $i$ people in the group of $58$ that have a coin in their right hand. This means there are $25-i$ people in the group of $25$ who have it in their right hand. So once you tell the people in the group of $25$ to switch hands, there are $i$ people in each group that have the coin in their right hand.

So, in the end, you don't know how many will have a coin in their right hand, but you know there are an equal number in each group.

$25 + 25 = \boxed{50}$

Now, to explain why this is the only solution...

Here is my start....

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Lets suppose you ask $n$ people to go in one group, then there are $83-n$ in the other.

Let $r$ be the number of "people with coins in their right hand" that are in the group of size $n$ . So, the group of size $83-n$ has $25-r$ "righties".

So, now you have:

Group	# of People	Coins in Right hand
$1$	$n$	$r$
$2$	$83-n$	$25-r$

Gotta run, I will plan to finish this off in a bit...

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Stay tuned...

Can you clarify what degree of freedom we have? IE Do we get to decide who goes into which group? Do we get to decide who switches their hand?

If yes, then I would move 12 right-hand people into the small group, and tell 1 right-hand person in the large group to switch his hand.

Thus, it sounds like there needs to be some restricted choice.

Calvin Lin Staff - 4 years, 2 months ago

OK, lemme try to clarify the question.

Geoff Pilling - 4 years, 2 months ago

There I've added a clarifying remark. Do you think it is sufficient?

Geoff Pilling - 4 years, 2 months ago

@Geoff Pilling - Your problem reminded me of this. :)

Anirudh Sreekumar - 4 years, 2 months ago

Haha... Yup, pretty similar! :)

Geoff Pilling - 4 years, 2 months ago

Hm, we could tell 58 people to switch the coins right?

Calvin Lin Staff - 4 years, 3 months ago

If we told 58 people to switch, then we aren't guaranteed to have equal numbers of coins in right hands. e.g. All of the original 25 "right handers" might be in the group of 25. Then when the 58 "lefties" switch it would leave 58 "right handers" in one group and 25 in the other.

Geoff Pilling - 4 years, 3 months ago

Ah yes. I just wrote out the algebra. Telling 58 people to switch coins would give us that those with coins in their left hand would be the same.

Calvin Lin Staff - 4 years, 3 months ago

83 beach goers

The answer is 50.

1 solution

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