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Old 2017-01-23, 21:29   #1
dwarfvader
 
Jan 2017

2 Posts
Default sum and product - advanced

Hope this is the right place for a puzzle, maybe you enjoy.

Possibly an unknown version of the impossible puzzle.

Let a and b be integers in the range of 2 to 1000 and a>=b.
Seth got the sum, Paul got the product and Quinn the quotient.

Paul: I do not know the two numbers.
Seth: I knew that.
Quinn: I've got a whole number.
Paul: Still don't know them.
Seth: Don't know the exact numbers yet.
Quinn: No clue. Sigh!
Seth: Now, I know them!
Paul: Didn't know them until Seth's statement, but now i've got them too.
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Old 2017-01-23, 22:03   #2
science_man_88
 
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"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
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Quote:
Originally Posted by dwarfvader View Post
Hope this is the right place for a puzzle, maybe you enjoy.

Possibly an unknown version of the impossible puzzle.

Let a and b be integers in the range of 2 to 1000 and a>=b.
Seth got the sum, Paul got the product and Quinn the quotient.

Paul: I do not know the two numbers.
Seth: I knew that.
Quinn: I've got a whole number.
Paul: Still don't know them.
Seth: Don't know the exact numbers yet.
Quinn: No clue. Sigh!
Seth: Now, I know them!
Paul: Didn't know them until Seth's statement, but now i've got them too.
Puzzles is also a good place

Paul's first comment says he doesn't know so there are multiple ways to get the product so a!=b except if they are a composite number. The fact that Quinn got a whole number shows that b divides into a therefore paul has a product x*b^2 this helps in the sense it deletes products that pass a squarefree test ( aka the distinct semiprimes etc.) edit: Seth's later statement about not knowing the exact numbers yet could hint that the sum has multiple factors as does x*b+b = (x+1)*(b) we could conclude that x+1 isn't a perfect power of b from this perhaps. or that b is not a perfect power of (x+1) also the fact that Quinn doesn't know shows that x is not unique to a and b ( could rule out quite a few values of x=1000/2=500 for example) it tells us a range for x.anyways for now I might stop to think about it more instead of editing my post

Last fiddled with by science_man_88 on 2017-01-23 at 22:24
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Old 2017-01-24, 21:20   #3
dwarfvader
 
Jan 2017

210 Posts
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You are right. Could a moderator please move this thread?
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