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Ray J. William Ray Norwood Jr. (born January 17, 1981), [1] known professionally as Ray J, is an American R&B singer, songwriter, television personality, and actor. Born in McComb, Mississippi, and raised in Carson, California, he is the younger brother of singer and actress Brandy Norwood. [3] In 2005, Ray J scored the Top 20 Billboard Hot 100 ...
In more generality: For all p ≥ 1 and odd h, f p − 1 (2 p h − 1) = 2 × 3 p − 1 h − 1. (Here f p − 1 is function iteration notation.) For all odd h, f(2h − 1) ≤ 3h − 1 / 2 The Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer n ≥ 1 such that f n (k) = 1. Undecidable ...
The probability of drawing another gold coin from the same box is 0 in (a), and 1 in (b) and (c). Thus, the overall probability of drawing a gold coin in the second draw is 0 / 3 + 1 / 3 + 1 / 3 = 2 / 3 . The problem can be reframed by describing the boxes as each having one drawer on each of two sides. Each ...
For instance, for Alcuin's version of the problem, =: a camel can carry 30 measures of grain and can travel one leuca while eating a single measure, where a leuca is a unit of distance roughly equal to 2.3 kilometres (1.4 mi). The problem has =: there are 90 measures of grain, enough to supply three trips.
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The 1/3–2/3 conjecture was formulated by Sergey Kislitsyn in 1968, [5] and later made independently by Michael Fredman [6] and by Nati Linial in 1984. [3] It was listed as a featured unsolved problem at the founding of the journal Order, and remains unsolved; being called "one of the most intriguing problems in the combinatorial theory of ...
The problem may be solved using simple addition. With 64 squares on a chessboard, if the number of grains doubles on successive squares, then the sum of grains on all 64 squares is: 1 + 2 + 4 + 8 + ... and so forth for the 64 squares. The total number of grains can be shown to be 2 64 −1 or 18,446,744,073,709,551,615 (eighteen quintillion ...
Problems 1–6 compute divisions of a certain number of loaves of bread by 10 men and record the outcome in unit fractions. Problems 7–20 show how to multiply the expressions 1 + 1/2 + 1/4 = 7/4, and 1 + 2/3 + 1/3 = 2 by different fractions. Problems 21–23 are problems in completion, which in modern notation are simply subtraction problems.