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In arbitrary-precision arithmetic, it is common to use long multiplication with the base set to 2 w, where w is the number of bits in a word, for multiplying relatively small numbers. To multiply two numbers with n digits using this method, one needs about n 2 operations.
The standard procedure for multiplication of two n-digit numbers requires a number of elementary operations proportional to , or () in big-O notation. Andrey Kolmogorov conjectured that the traditional algorithm was asymptotically optimal , meaning that any algorithm for that task would require Ω ( n 2 ) {\displaystyle \Omega (n^{2 ...
Another method is to simply multiply the number by 10, and add the original number to the result. For example: 17 × 11 = ? 17 × 10 = 170 170 + 17 = 187 17 × 11 = 187 One last easy way: If one has a two-digit number, take it and add the two numbers together and put that sum in the middle, and one can get the answer.
Trachtenberg defined this algorithm with a kind of pairwise multiplication where two digits are multiplied by one digit, essentially only keeping the middle digit of the result. By performing the above algorithm with this pairwise multiplication, even fewer temporary results need to be held. Example:
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Many common methods for multiplying numbers using pencil and paper require a multiplication table of memorized or consulted products of small numbers (typically any two numbers from 0 to 9). However, one method, the peasant multiplication algorithm, does not. The example below illustrates "long multiplication" (the "standard algorithm", "grade ...
If the sum contains more than one digit, the value of the tens place is carried into the next diagonal (see Step 2). Step 2. Numbers are filled to the left and to the bottom of the grid, and the answer is the numbers read off down (on the left) and across (on the bottom). In the example shown, the result of the multiplication of 58 with 213 is ...
Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...