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First multiply the quarters by 47, the result 94 is written into the first workspace. Next, multiply cwt 12*47 = (2 + 10)*47 but don't add up the partial results (94, 470) yet. Likewise multiply 23 by 47 yielding (141, 940). The quarters column is totaled and the result placed in the second workspace (a trivial move in this case).
Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y. One must multiply the leftmost digit of the original number by 3, add the next digit, take the remainder when divided by 7, and continue from the beginning: multiply by 3, add the next digit, etc.
so 3 × 17 = 30 + 21 = 51. This is the "grid" or "boxes" structure which gives the multiplication method its name. Faced with a slightly larger multiplication, such as 34 × 13, pupils may initially be encouraged to also break this into tens. So, expanding 34 as 10 + 10 + 10 + 4 and 13 as 10 + 3, the product 34 × 13 might be represented:
The method for general multiplication is a method to achieve multiplications with low space complexity, i.e. as few temporary results as possible to be kept in memory. . This is achieved by noting that the final digit is completely determined by multiplying the last digit of the multiplic
[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.
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. These are all added to produce 17. The units digit of this, 7, is written as the next digit of the result.
Here, 29 is greater than 1 cubed, greater than 2 cubed, greater than 3 cubed, but not greater than 4 cubed. The greatest cube it is greater than is 3, so the first digit of the two-digit cube must be 3. Therefore, the cube root of 29791 is 31. Another example: Find the cube root of 456533. The cube root ends in 7.
Dividing 272 and 8, starting with the hundreds digit, 2 is not divisible by 8. Add 20 and 7 to get 27. The largest number that the divisor of 8 can be multiplied by without exceeding 27 is 3, so it is written under the tens column. Subtracting 24 (the product of 3 and 8) from 27 gives 3 as the remainder.
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