\Theta } (N), but usually it will go much faster. For multiplication, the most straightforward algorithms used for multiplying numbers by hand (as taught in primary school) require Θ {\displaystyle \Theta } (N2) operations, but multiplication algorithms that achieve O(N log(N) log(log(N))) complexity have been devised, such as the Schönhage–Strassen algorithm, based on fast Fourier transforms, and there are also algorithms with slightly worse complexity but with sometimes superior real-world performance for smaller N. The Karatsuba multiplication is such an algorithm. For division, see division algorithm. For a list of algorithms along with complexity estimates, see computational complexity of mathematical operations. For examples in x86 assembly, see external links. In some languages such as REXX, the precision of all calculations must be set before doing a calculation. Other languages, such as Python and Ruby extend the precision automatically to prevent overflow. The calculation of factorials can easily produce very large numbers. This is not a problem for their usage in many formulae (such as Taylor series) because they appear along with other terms, so that—given careful attention to the order of evaluation—intermediate calculation values are not troublesome. If approximate values of factorial numbers are desired, Stirling's approximation gives good results using floating-point arithmetic. The largest representable value for a fixed-size integer variable may be exceeded even for relatively small arguments as shown in the table below. Even floating-point numbers are soon outranged, so it may help to recast the calculations in terms of the logarithm of the number. But if exact values for large factorials are desired, then special software is required, as in the pseudocode that follows, which implements the classic algorithm to calculate 1, 1×2, 1×2×3, 1×2×3×4, etc. the successive factorial numbers. Constant Limit = 1000; % Sufficient digits. Constant Base = 10; % The base of the simulated arithmetic. Constant FactorialLimit = 365; % Target number to solve, 365! Array digit[1:Limit] of integer; % The big number. Integer carry,d; % Assistants during multiplication. Integer last,i; % Indices to the big number's digits. Array text[1:Limit] of character; % Scratchpad for the output. Constant tdigit[0:9] of character = ["0","1","2","3","4","5","6","7","8","9"]; BEGIN digit:=0; % Clear the whole array. digit[1]:=1; % The big number starts with 1, last:=1; % Its highest-order digit is number 1. for n:=1 to FactorialLimit do % Step
81 and 162 are both multiples of 3. In mathematics, a number that can be divided by another number without leaving a remainder is said to be divisible by that number. Both 81 and 162 can be evenly divided by 3, making them multiples of 3.
5
The hcf of 45 and 40 is 5.
It is: 9
It is: 27
hcf(18, 54, 81) = 9. Using prime factorization: 18 = 2 x 3^2 54 = 2 x 3^3 81 = 3^4 hcf = 3^2 = 9
The HCF is 1.
The GCF is 1.
The GCF is 27.
HCF(26, 45) = 1. The numbers are co-prime.
The GCF is 15.
To find the highest common factor (HCF) of 81, 243, and 1215, we first need to find the prime factorization of each number. The prime factorization of 81 is 3^4, 243 is 3^5, and 1215 is 3^5 * 5. To find the HCF, we look for the highest power of each prime factor that appears in all the numbers. In this case, the highest power of 3 that appears in all three numbers is 3^4, so the HCF of 81, 243, and 1215 is 81.