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The base 3 appears 5 times in the multiplication, because the exponent is 5. Here, 243 is the 5th power of 3, or 3 raised to the 5th power. The word "raised" is usually omitted, and sometimes "power" as well, so 3 5 can be simply read "3 to the 5th", or "3 to the 5".
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 ...
A power of two is a number of the form 2n where n is an integer, that is, the result of exponentiation with number two as the base and integer n as the exponent . Powers of two with non-negative exponents are integers: 20 = 1, 21 = 2, and 2n is two multiplied by itself n times. [ 1][ 2] The first ten powers of 2 for non-negative values of n are:
Fourth power. In arithmetic and algebra, the fourth power of a number n is the result of multiplying four instances of n together. So: n4 = n × n × n × n. Fourth powers are also formed by multiplying a number by its cube. Furthermore, they are squares of squares. Some people refer to n4 as n “ tesseracted ”, “ hypercubed ...
The term hyperpower [4] is a natural combination of hyper and power, which aptly describes tetration. The problem lies in the meaning of hyper with respect to the hyperoperation sequence. When considering hyperoperations, the term hyper refers to all ranks, and the term super refers to rank 4, or tetration.
Problem II.8 of the Arithmetica asks how a given square number is split into two other squares; in other words, for a given rational number k, find rational numbers u and v such that k 2 = u 2 + v 2. Diophantus shows how to solve this sum-of-squares problem for k = 4 (the solutions being u = 16/5 and v = 12/5 ).
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