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If the top number is too small to subtract the bottom number from it, we add 10 to it; this 10 is "borrowed" from the top digit to the left, which we subtract 1 from. Then we move on to subtracting the next digit and borrowing as needed, until every digit has been subtracted. Example: [citation needed]
110 ÷ 5 = 22 (The result is the same as the original number divided by 5) If the last digit is 5. 85 (The original number) 8 5 (Take the last digit of the number, and check if it is 0 or 5) 8 5 (If it is 5, take the remaining digits, discarding the last) 8 × 2 = 16 (Multiply the result by 2) 16 + 1 = 17 (Add 1 to the result)
The black numbers are the addends, the green number is the carry, and the blue number is the sum. In the rightmost digit, the addition of 9 and 7 is 16, carrying 1 into the next pair of the digit to the left, making its addition 1 + 5 + 2 = 8. Therefore, 59 + 27 = 86.
If this number is truncated to 4 decimal places, the result is 3.141. Rounding is a similar process in which the last preserved digit is increased by one if the next digit is 5 or greater but remains the same if the next digit is less than 5, so that the rounded number is the best approximation of a given precision for the original number.
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 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. For example: 24 x 11 = 264 because 2 + 4 = 6 and the 6 is placed in between the 2 and the 4. Second example: 87 x 11 = 957 because 8 + 7 = 15 so the 5 goes in between the 8 and the 7 and the 1 is carried to the 8.
The exception occurs when the original number has a digital root of 9, whose digit sum is itself, and therefore will not be cast out by taking further digit sums. The number 12565, for instance, has digit sum 1+2+5+6+5 = 19, which, in turn, has digit sum 1+9=10, which, in its turn has digit sum 1+0=1, a single-digit number.
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.
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