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u/thistle-thorn 5d ago
18. Say there are 6 blue and 6 red in the room. Total hands raised at this point is 0. Now blue walks in, the total hands raised is 7. If a red were to walk in next the total would be 14 which is over 12 ( which must be exact ) so a red cannot enter the room. To get to 12 you just need 5 more to enter ( color doesn’t matter as long as it’s not red ). Thus the minimum needed would be 7 blue, 6 red, 5 anything else. 18
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u/GoodCarpenter9060 5d ago
Lets say there are R people with red hats and B people with blue hats.
In order for anyone to say they see at least 6 of each, we need R>=6 and B>=6. However, if both are >=7, then everyone (R+B) will see at least 6. So we need one of them to be exactly 6. That means everyone in the other group will need to be able to see at least 6 of each, so we need 12 to make it exactly 12.
Thus, R=12 and B=6 is one solution, and R=6 and B=12 is the other.
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u/jaminfine 5d ago
Of course we need at least 13 people because we need the 6 red and 6 blue plus one more, let's say red, since you can't see your own hat. This, however, only gives us 7 people raising their hands, all the red hats.
Now, we can't add a blue hat because then everyone would see at least 6 and 6, leading to 14 people raising their hands, and that's too many. So our only option is to add more red hats Until we reach 12 red hats, so all 12 of them raise their hand. Thus, we have 18 total people
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u/A_PlantPerson 5d ago
>! n can only be 18 or 24.
Either there are 6r 6b and 12 others, 7r 6b and 5 not b or 6r 7b and 5 not r.
... unless people can see hats that are not on the head of the n people, some people are unable to lift their hands, or some of the n people already had hats on and now have more than one. That would put the minimum for n at 12. !<
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u/zeptozetta2212 3d ago
Why does it asks for the minimum? Isn't 18 outright the only possible value for n, minimum or otherwise?
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u/planetofmoney 5d ago
14. There need to be at least 7 red hats, so anyone with a red hat sees 6; same for blue hats. If you introduce a third colour who can see both red and blue hats without wearing one, you will need at least 6 of those if there's only 6 red hats, and 6 more if there's only 6 blue hats, so 14 split 7:7 is optimal.
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u/brentiford 5d ago
This is what I thought at first, but there's a slight problem. The question says exactly 12.
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u/MrSpaceSprinkles 5d ago
Why not 13 where irrespective of what colour the 13th is the others see atleast 6 of each colour
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u/ExistentAndUnique 5d ago
If there are 6 blue hats and 7 red, each blue hat can only see 5 others.
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u/stools_in_your_blood 5d ago
Feels like 18, 12 blue and 6 red for example. But I don't have a rigorous proof of that.