Puzzle Quiz
- Thread starter Renata
- Start date
Perfection
The Great Head.
2 3 & 6
Pontiuth Pilate
Republican Jesus!
Er, I'm not sure how Perfection got that answer?
An explanation would be nice
An explanation would be nice
Because 2, 3, and 6 are the only three numbers that form 36 without the two smaller numbers being the same. (After all, there has to be a youngest son.)Originally posted by Pontiuth Pilate
Er, I'm not sure how Perfection got that answer?
An explanation would be nice
Syterion
Voodoo Economist
Look at the details more, WillJ.
Perfection
The Great Head.
Actually 1 2 and 18 should be equally valid
Pontiuth Pilate
Republican Jesus!
Ah. Smart
Explanation for the Pirate Problem using reverse engineering:
A proposes: he gets all 100
B proposes: no matter what B proposes, A will reject so that he gets the entire treasure chest.
C proposes: A will always reject C so that he can then kill B and get it all. B will always accept C so that B at least gets to live. C can therefore propose [0, 0, 100] which will always pass.
D proposes: C will always reject D so he can propose his own deal. D needs 3 votes [himself, A, and B]. To get their votes he needs to give them more gold than they'd get from C. Therefore
D proposes [1, 1, 0, 98] which passes - because A and B get nothing if C gets to propose [since B will always vote for 0-0-100 so he at least gets to live], so A and B settle for 1 coin each.
That far I'm sure of - now what about E?
Explanation for the Pirate Problem using reverse engineering:
A proposes: he gets all 100
B proposes: no matter what B proposes, A will reject so that he gets the entire treasure chest.
C proposes: A will always reject C so that he can then kill B and get it all. B will always accept C so that B at least gets to live. C can therefore propose [0, 0, 100] which will always pass.
D proposes: C will always reject D so he can propose his own deal. D needs 3 votes [himself, A, and B]. To get their votes he needs to give them more gold than they'd get from C. Therefore
D proposes [1, 1, 0, 98] which passes - because A and B get nothing if C gets to propose [since B will always vote for 0-0-100 so he at least gets to live], so A and B settle for 1 coin each.
That far I'm sure of - now what about E?
Perfection
The Great Head.
In fact 3,3, and 4 is correct because there still would be a younger sibling (even if we disinclude twins we could have kids born 11 months apart
I'm just going to go and do my own, because damn it I deserve it
I'm just going to go and do my own, because damn it I deserve it
Syterion
Voodoo Economist
Nobody has gotten it yet. I don't care if they could be 11 months apart, assume they are twins who were born at the exact same second if the integer representation of their age is the same. Here's a hint: Try to think of what Tom is thinking of during the course of the dialogue.
Perfection
The Great Head.
Edit: silly thought
Perfection
The Great Head.
perhaps it's just 2,3,6 because 1 year olds tend not to have hair
No, it's definitely 1, 6, and 6. For a demonstration, here is the list of the three-way factorizations of 36:
1 1 36
1 2 18
1 3 12
1 4 9
1 6 6
2 2 9
2 3 6
3 3 4
All add up to different sums, except for 1 6 6 and 2 2 9, which both add up to 13 and therefore Tom can't know which one is true. (Obviously the current date is the 13th.) And then Tim mentions a youngest son, which means the answer has to be 1 6 6.
Syterion
Voodoo Economist
Good job WillJ. It was just a coincidence that today happened to be the 13th of March as well.
Perfection
The Great Head.
Fine Go ahead, steal my spot even though I got the last 2 and didn't get to put down the awesome puzzle I just came up with
Syterion
Voodoo Economist
Post it anyway. I never got to actually solve one.
Perfection
The Great Head.
Okay, here we go... 
THE INDESTRUCTABLE SQUARE OF DOOM is a square piece of paper ten inches by ten inches.
If you place two distinct points on it so that all points are equidistant the farthest they can be from each other is 10sqrt(2)
If you place three distinct points on it so that all points are equidistant the farthest they can be from each other is 10(sqrt(3)-1)2
If you place four distinct points on it so that all points are equidistant what is the farthest they can be from each other?
THE INDESTRUCTABLE SQUARE OF DOOM is a square piece of paper ten inches by ten inches.
If you place two distinct points on it so that all points are equidistant the farthest they can be from each other is 10sqrt(2)
If you place three distinct points on it so that all points are equidistant the farthest they can be from each other is 10(sqrt(3)-1)2
If you place four distinct points on it so that all points are equidistant what is the farthest they can be from each other?
