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For example, 3 5 = 3 · 3 · 3 · 3 · 3 = 243. 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".
In arithmetic and algebra, the seventh power of a number n is the result of multiplying seven instances of n together. So: n 7 = n × n × n × n × n × n × n. Seventh powers are also formed by multiplying a number by its sixth power, the square of a number by its fifth power, or the cube of a number by its fourth power.
In arithmetic and algebra, the fifth power or sursolid [1] of a number n is the result of multiplying five instances of n together: n 5 = n × n × n × n × n. Fifth powers are also formed by multiplying a number by its fourth power, or the square of a number by its cube. The sequence of fifth powers of integers is:
y = x 3 for values of 1 ≤ x ≤ 25.. In arithmetic and algebra, the cube of a number n is its third power, that is, the result of multiplying three instances of n together. The cube of a number n is denoted n 3, using a superscript 3, [a] for example 2 3 = 8.
The sequence starts with a unary operation (the successor function with n = 0), and continues with the binary operations of addition (n = 1), multiplication (n = 2), exponentiation (n = 3), tetration (n = 4), pentation (n = 5), etc. Various notations have been used to represent hyperoperations.
In 1992, Roger Heath-Brown conjectured that every unequal to 4 or 5 modulo 9 has infinitely many representations as sums of three cubes. [56] The case n = 33 {\displaystyle n=33} of this problem was used by Bjorn Poonen as the opening example in a survey on undecidable problems in number theory , of which Hilbert's tenth problem is the most ...
To compute the largest power of 2 dividing the binomial coefficient () write m = 3 and n − m = 7 in base p = 2 as 3 = 11 2 and 7 = 111 2.Carrying out the addition 11 2 + 111 2 = 1010 2 in base 2 requires three carries:
2 = 3 3 − 5 2 10 = 13 3 − 3 7 18 = 19 2 − 7 3 = 3 5 − 15 2. It had been conjectured that 6 cannot be so represented, and Golomb conjectured that there are infinitely many integers which cannot be represented as a difference between two powerful numbers. However, Narkiewicz showed that 6 can be so represented in infinitely many ways such as