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Both these methods break up the subtraction as a process of one digit subtractions by place value. Starting with a least significant digit, a subtraction of the subtrahend: s j s j−1... s 1. from the minuend m k m k−1... m 1, where each s i and m i is a digit, proceeds by writing down m 1 − s 1, m 2 − s 2, and so forth, as long as s i ...
For example, through the standard addition algorithm, the sum can be obtained by following three rules: a) line up the digits of each addend by place value, longer digit addends should go on top, b) each addend can be decomposed -- ones are added with ones, tens are added with tens, and so on, and c) if the sum of the digits of the current place value is ten or greater, then the number must be ...
Here, 7 − 9 = −2, so try (10 − 9) + 7 = 8, and the 10 is got by taking ("borrowing") 1 from the next digit to the left. There are two ways in which this is commonly taught: The ten is moved from the next digit left, leaving in this example 3 − 1 in the tens column.
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.
The nines' complement of a decimal digit is the number that must be added to it to produce 9; the nines' complement of 3 is 6, the nines' complement of 7 is 2, and so on, see table. To form the nines' complement of a larger number, each digit is replaced by its nines' complement. Consider the following subtraction problem:
If we were to express this idea using symbols of grouping, the factors in a product. Example: 2+3×4 = 2 +(3×4)=2+12=14. In understanding expressions without symbols of grouping, it is useful to think of subtraction as addition of the opposite, and to think of division as multiplication by the reciprocal.
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