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Conversely the period of the repeating decimal of a fraction c / d will be (at most) the smallest number n such that 10 n − 1 is divisible by d. For example, the fraction 2 / 7 has d = 7, and the smallest k that makes 10 k − 1 divisible by 7 is k = 6, because 999999 = 7 × 142857. The period of the fraction 2 / 7 is ...
In some contexts it is desirable to round a given number x to a "neat" fraction – that is, the nearest fraction y = m/n whose numerator m and denominator n do not exceed a given maximum. This problem is fairly distinct from that of rounding a value to a fixed number of decimal or binary digits, or to a multiple of a given unit m .
The base determines the fractions that can be represented; for instance, 1/5 cannot be represented exactly as a floating-point number using a binary base, but 1/5 can be represented exactly using a decimal base (0.2, or 2 × 10 −1).
The set of rational numbers includes all integers, which are fractions with a denominator of 1. The symbol of the rational numbers is Q {\displaystyle \mathbb {Q} } . [ 19 ] Decimal fractions like 0.3 and 25.12 are a special type of rational numbers since their denominator is a power of 10.
For example, the set of integers modulo 12 has twelve elements; it inherits an addition operation from the integers that is central to musical set theory. The set of integers modulo 2 has just two elements; the addition operation it inherits is known in Boolean logic as the "exclusive or" function.
As far as is known, this is not possible using classical (non-quantum) computers; no classical algorithm is known that can factor integers in polynomial time. However, Shor's algorithm shows that factoring integers is efficient on an ideal quantum computer, so it may be feasible to defeat RSA by constructing a large quantum computer.
More formally, given sets A and S, and some sets U and V of surjections A → S, we often wish to know the number of pairs of functions (f, g) such that f is in U and g is in V, and for all a in A, f(a) ≠ g(a); in other words, where for each f and g, there exists a derangement φ of S such that f(a) = φ(g(a)).
Multiply together the results of the two previous steps The product of all primes up to n {\displaystyle n} is an O ( n ) {\displaystyle O(n)} -bit number, by the prime number theorem , so the time for the first step is O ( n log 2 n ) {\displaystyle O(n\log ^{2}n)} , with one logarithm coming from the divide and conquer and another coming ...