A Logic Party
I organized a party which had 100 guests (including me). Several of us shook each other's hands (once), and we all shook at least one person's hand.
When we lined up, we found out that the first guest shook 1 hand, the second guest shook 2 hands, the third guest shook 3 hands, so on and so forth, with the ninety-nineth guess shaking 99 hands.
How many hands did the one hundredth guess shake?
The answer is 50.
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1 solution
(1) Let's denote n S j if the guest # n greet the guest# j . Now n S j ≡ j S n because nobody can greet to another with a clashing hand if the another don't clashes her hand too.
(2) As 99 S 1 ⟹ 1 S 99 ⟹ 1 give all his greetings. And 99 greeting at all in the party 99 S 2, 99 S 3..., 99 S 98, 99 S 100 ⟹ 100 have 1 greet.
(3) 98 S 1 ⟹ 98 S 2, 98 S 3..., 98 S 97, 98 S 99, 98 S 100 and complete his 98 greetings and #2 give all his greetings
(3) We can repeat this process and it implies that 51 S 49, 51 S 52,..., 51 S 100. At this point 100 give 49 greetings.
(4) Now 50 S 51, 50 S 51,..., 50 S 100 and #50 give all his 50 greetings. And this point #100 give 50 greetings.
(5) And here finish the greetings beacuse like be mutuals, continue with #49 implies a double greetings.
Then #100 give 50 grettings