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        riddles >> hard >> Another variation to the sum / product riddle
        
(Message started by: Altamira_64 on Feb 1^st, 2013, 6:17am)

Title: Another variation to the sum / product riddle
Post by Altamira_64 on Feb 1^st, 2013, 6:17am

3 logicians A, P and S are again bored to death and looking for ways to drive us crazy :)

A thinks of two integers >1 and with sum <100.
He then announces their product to P and their sum to S.

The following conversation occurs.
P says "I cannot find these numbers."
S says "I was sure that you could not find them."
P says "I knew in advance that you knew I could not find them"
S says "I don't know the numbers."
P says "Now I know them!"

What are these numbers?

Title: Re: Another variation to the sum / product riddle
Post by SMQ on Feb 2^nd, 2013, 6:32am

P says "I cannot find these numbers." --> [hide]the product is not unique.[/hide]

S says "I was sure..." --> [hide]all pairs with this sum have non-unique products.[/hide]

P says "I knew in advance..." --> [hide]all pairs with this product have a sum such that (all pairs with this sum have non-unique products).[/hide]

[hide]By exhaustive search in excel, this leaves 7 candidate pairs:
6, 17 (sum 23, product 102)
13, 22 (sum 35, product 286)
3, 34 (sum 37, product 102)
11, 26 (sum 37, product 286)
14, 23 (sum 37, product 322)
2, 51 (sum 53, product 102)
7, 46 (sum 53, product 322)
[/hide]
S says "I don't know the numbers." --> [hide]the sum is not unique among the candidate pairs[/hide]

[hide]This leaves 5 remaining candidates:
3, 34 (sum 37, product 102)
2, 51 (sum 53, product 102)
11, 26 (sum 37, product 286)
14, 23 (sum 37, product 322)
7, 46 (sum 53, product 322)
[/hide]
P says: "Now I know them!" --> [hide]the product is unique among the remaining candidates[/hide]

[hide]The only remaining possibility is the pair 11, 26[/hide]

--SMQ