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The Basel problem is a problem in mathematical analysis with relevance to number theory, concerning an infinite sum of inverse squares. It was first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734, [1] and read on 5 December 1735 in The Saint Petersburg Academy of Sciences. [2] Since the problem had withstood the attacks of ...
For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 / 2 ≡ 2 (mod 3). Equivalently, 2n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 2 (mod 3). Conjecturally, this inverse relation forms a tree except for a 1–2 loop (the inverse of the 1–2 loop of the function f(n) revised as indicated above).
It is now clear that if the warden answers B to A ( 1 / 2 of all cases), then 1 / 3 of the time C is pardoned and A will still be executed (case 4), and only 1 / 6 of the time A is pardoned (case 1). Hence C's chances are ( 1 / 3 )/( 1 / 2 ) = 2 / 3 and A's are ( 1 / 6 )/( 1 / 2 ) = 1 / ...
The misdirection in this riddle is in the second half of the description, where unrelated amounts are added together and the person to whom the riddle is posed assumes those amounts should add up to 30, and is then surprised when they do not — there is, in fact, no reason why the (10 − 1) × 3 + 2 = 29 sum should add up to 30.
The Hardest Logic Puzzle Ever is a logic puzzle so called by American philosopher and logician George Boolos and published in The Harvard Review of Philosophy in 1996. [1][2] Boolos' article includes multiple ways of solving the problem. A translation in Italian was published earlier in the newspaper La Repubblica, under the title L'indovinello ...
To fully solve the problem, a simple tree is formed with the initial state as the root. The five possible actions ( 1,0,1 , 2,0,1 , 0,1,1 , 0,2,1 , and 1,1,1 ) are then subtracted from the initial state, with the result forming children nodes of the root. Any node that has more cannibals than missionaries on either bank is in an invalid state ...
A veridical paradox is a paradox whose correct solution seems to be counterintuitive. It may seem intuitive that the probability that the remaining coin is gold should be 1 / 2 , but the probability is actually 2 / 3 . [1] Bertrand showed that if 1 / 2 were correct, it would result in a contradiction, so 1 / 2 ...
Monty Hall problem. In search of a new car, the player chooses a door, say 1. The game host then opens one of the other doors, say 3, to reveal a goat and offers to let the player switch from door 1 to door 2. The Monty Hall problem is a brain teaser, in the form of a probability puzzle, based nominally on the American television game show Let ...