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[1] [2] [3] It is a divide-and-conquer algorithm that reduces the multiplication of two n-digit numbers to three multiplications of n/2-digit numbers and, by repeating this reduction, to at most single-digit multiplications.
To multiply two numbers with n digits using this method, one needs about n 2 operations. More formally, multiplying two n-digit numbers using long multiplication requires Θ(n 2) single-digit operations (additions and multiplications).
In this case, the column digit is 5 and the row digit is 2. Write their product, 10, in the cell, with the digit 1 above the diagonal and the digit 0 below the diagonal (see picture for Step 1). If the simple product lacks a digit in the tens place, simply fill in the tens place with a 0. [2] Step 1
The next band from the right has three digits, 2, 1 and 8. These are added together to get 11. The units digit of this addition, 1, is written down as the next digit of the multiplication result. The tens digit, which is 1, is carried into the next band. The third band from the right has five digits, 2, 4, 3, 1 and 6 plus the carried 1.
[2] [3] If the last digit in the number is 0, then the result will be the remaining digits multiplied by 2. For example, the number 40 ends in a zero, so take the remaining digits (4) and multiply that by two (4 × 2 = 8). The result is the same as the result of 40 divided by 5(40/5 = 8).
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:
For instance, the product of three factors of two (2×2×2) is "two raised to the third power", and is denoted by 2 3, a two with a superscript three. In this example, the number two is the base, and three is the exponent. [26] In general, the exponent (or superscript) indicates how many times the base appears in the expression, so that the ...
The last digit is the check digit "7", and if the other numbers are correct then the check digit calculation must produce 7. Add the odd number digits: 0+6+0+2+1+5 = 14. Multiply the result by 3: 14 × 3 = 42. Add the even number digits: 3+0+0+4+4 = 11. Add the two results together: 42 + 11 = 53.
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