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Old 07-18-2022, 08:03 AM   #41
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I hope these have been fun problems. I've had several in my head from when I first heard them, but I'm starting to look around for more - hopefully I can keep finding interesting ones to solve that don't require lots of background knowledge. Again, if anyone else has any to share, please do.

Here's a new problem. Let's see if it gets solved before the weekend is over or not.

You have a complete graph on 6 vertices (graph where all vertices are connected to each other) like below


Each edge can be colored either red or blue.

Show that regardless of the coloring, you can always find 3 edges that form a triangle that will have its edges all the same color.
No takers on this one?

The proof is pretty (imho):

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Old 07-18-2022, 09:08 AM   #42
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Ok, new puzzle. This one is a pretty cool one I remember from my high school math teacher.

You have 100 coins on a table, with 50 heads, 50 tails. You are blindfolded and must divide the coins into two groups of 50 such that each group will have the same number of heads/tails.

The coins are each encased in plastic so you can't tell whether a coin is heads or tails by touch.

How do you do this?
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Old 07-18-2022, 07:55 PM   #43
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Ok, new puzzle. This one is a pretty cool one I remember from my high school math teacher.

You have 100 coins on a table, with 50 heads, 50 tails. You are blindfolded and must divide the coins into two groups of 50 such that each group will have the same number of heads/tails.

The coins are each encased in plastic so you can't tell whether a coin is heads or tails by touch.

How do you do this?
I wish I knew!
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Old 07-18-2022, 10:21 PM   #44
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Ok, new puzzle. This one is a pretty cool one I remember from my high school math teacher.

You have 100 coins on a table, with 50 heads, 50 tails. You are blindfolded and must divide the coins into two groups of 50 such that each group will have the same number of heads/tails.

The coins are each encased in plastic so you can't tell whether a coin is heads or tails by touch.

How do you do this?
To confirm this isn’t a cheesy answer like each coin has a head and a tail therefore I split them into to piles of 50 correct?
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Old 07-18-2022, 10:26 PM   #45
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To confirm this isn’t a cheesy answer like each coin has a head and a tail therefore I split them into to piles of 50 correct?
Correct, it's not a trick question.
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Old 07-18-2022, 11:26 PM   #46
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Old 07-18-2022, 11:43 PM   #47
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Originally Posted by psyang View Post
Ok, new puzzle. This one is a pretty cool one I remember from my high school math teacher.

You have 100 coins on a table, with 50 heads, 50 tails. You are blindfolded and must divide the coins into two groups of 50 such that each group will have the same number of heads/tails.

The coins are each encased in plastic so you can't tell whether a coin is heads or tails by touch.

How do you do this?
Not sure if GGG already gave this answer as I didn't look at his, but here goes...
Spoiler!
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Old 07-19-2022, 07:26 AM   #48
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Not sure if GGG already gave this answer as I didn't look at his, but here goes...
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Old 07-19-2022, 08:08 AM   #49
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I did think about that after I posted, but I still don't think it's a trick question. It is worded to say exactly what it is supposed to say and I don't know how to word the question differently without it being a hint.
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Old 07-19-2022, 08:17 AM   #50
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And we have two winners on this one! Great work to both of you.

Spoiler!


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Quote:
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Not sure if GGG already gave this answer as I didn't look at his, but here goes...
Spoiler!
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Old 07-19-2022, 08:41 AM   #51
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A quick follow up to the last one.

You have the same setup - 100 coins encased in plastic, you are blindfolded. Only this time, there are 10 tails. Can you divide the coins into two groups so that they each have the same number of tails?

And now, generalizing: What if there are n tails at the start. Can you always divide the coins into two groups so that they have the same number of tails?
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Old 07-19-2022, 08:57 AM   #52
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A quick follow up to the last one.

You have the same setup - 100 coins encased in plastic, you are blindfolded. Only this time, there are 10 tails. Can you divide the coins into two groups so that they each have the same number of tails?

And now, generalizing: What if there are n tails at the start. Can you always divide the coins into two groups so that they have the same number of tails?
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Old 07-19-2022, 10:36 AM   #53
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Bring_Back_Shantz is the winner - congrats! For the bonus point, you should have mentioned that n (or x in your solution) needs to be even, otherwise it is impossible.

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Old 07-19-2022, 10:42 AM   #54
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Ok, one more coin question (for now). This one might be tricky, we'll see. The solution is very elegant.

You have 100 perfectly round and identical quarters on a rectangular table arranged such that none of them overlap (ie. all quarters are flat on the table) and it is impossible to add another quarter to the table without requiring an overlap.

Show that you can always cover the table completely with 400 coins (where, by "cover", I mean that for every point on the table, there is at least one coin directly above it). Obviously, the covering would allow for overlaps.
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Old 07-19-2022, 10:50 AM   #55
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Bring_Back_Shantz is the winner - congrats! For the bonus point, you should have mentioned that n (or x in your solution) needs to be even, otherwise it is impossible.

Spoiler!
I assume you mean "T" in my solution.
As in the starting number of tails in the total group.
But that's not correct, it can definitely be odd.
Look at the extreme example where n = 1 (1 starting tail)

Group 1 will be 1 coin.
Group 2 will be 99 coins.

There are 2 possibilities
1:
Group 1 = 1 tail
Group 2 = 99 heads

Flip over the coin in group 1
Group 1 = 1 head = 0 tails
Group 2 = 99 heads = 0 tails

2:
Group 1 = 1 Head
Group 2 = 1 tail, 98 heads

Flip over the coin in group 1
Group 1 - 1 tail
Group 2 = 1 tail, 98 heads

Works just fine for an odd number.
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Old 07-19-2022, 10:57 AM   #56
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I assume you mean "T" in my solution.
As in the starting number of tails in the total group.
But that's not correct, it can definitely be odd.
Look at the extreme example where n = 1 (1 starting tail)

Group 1 will be 1 coin.
Group 2 will be 99 coins.

There are 2 possibilities
1:
Group 1 = 1 tail
Group 2 = 99 heads

Flip over the coin in group 1
Group 1 = 1 head = 0 tails
Group 2 = 99 heads = 0 tails

2:
Group 1 = 1 Head
Group 2 = 1 tail, 98 heads

Flip over the coin in group 1
Group 1 - 1 tail
Group 2 = 1 tail, 98 heads

Works just fine for an odd number.
Ha ha, you are right. I don't know what I was thinking! Well, I do know what I was thinking but it was obviously wrong. Thanks for the correction!
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Old 07-19-2022, 03:20 PM   #57
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Ok, one more coin question (for now). This one might be tricky, we'll see. The solution is very elegant.

You have 100 perfectly round and identical quarters on a rectangular table arranged such that none of them overlap (ie. all quarters are flat on the table) and it is impossible to add another quarter to the table without requiring an overlap.

Show that you can always cover the table completely with 400 coins (where, by "cover", I mean that for every point on the table, there is at least one coin directly above it). Obviously, the covering would allow for overlaps.

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Old 07-19-2022, 04:29 PM   #58
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Hmm. Not quite.

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Old 07-19-2022, 05:44 PM   #59
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Old 07-19-2022, 06:29 PM   #60
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I think it's a great start but not a valid solution. Comments/questions below in blue:

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