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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 () elementary operations.
More formally, multiplying two n-digit numbers using long multiplication requires Θ(n 2) single-digit operations (additions and multiplications). When implemented in software, long multiplication algorithms must deal with overflow during additions, which can be expensive.
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 ...
Two -digit numbers One +-digit ... One -digit number Schoolbook long multiplication () Karatsuba algorithm 3-way Toom–Cook multiplication ()-way Toom ...
A small number is chosen, usually 2 through 9, by which to multiply the large number. In this example the small number by which to multiply the larger is 6. The horizontal row in which this number stands is the only row needed to perform the remaining calculations and may now be viewed in isolation. Second step of solving 6 x 425
The sum of two numbers is unique; there is only one correct answer for a sums. [8] When the sum of a pair of digits results in a two-digit number, the "tens" digit is referred to as the "carry digit". [9] In elementary arithmetic, students typically learn to add whole numbers and may also learn about topics such as negative numbers and fractions.
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 ...