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If necessary, simplify the long division problem by moving the decimals of the divisor and dividend by the same number of decimal places, to the right (or to the left), so that the decimal of the divisor is to the right of the last digit. When doing long division, keep the numbers lined up straight from top to bottom under the tableau.
More systematically and more efficiently, two integers can be divided with pencil and paper with the method of short division, if the divisor is small, or long division, if the divisor is larger. If the dividend has a fractional part (expressed as a decimal fraction ), one can continue the procedure past the ones place as far as desired.
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
In computing, the modulo operation returns the remainder or signed remainder of a division, after one number is divided by another, called the modulus of the operation.. Given two positive numbers a and n, a modulo n (often abbreviated as a mod n) is the remainder of the Euclidean division of a by n, where a is the dividend and n is the divisor.
In the division of 43 by 5, we have: 43 = 8 × 5 + 3, so 3 is the least positive remainder. We also have that: 43 = 9 × 5 − 2, and −2 is the least absolute remainder. These definitions are also valid if d is negative, for example, in the division of 43 by −5, 43 = (−8) × (−5) + 3, and 3 is the least positive remainder, while,
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.
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) 85 ÷ 5 = 17 (The result is the same as the original number divided by 5)
As in all division problems, a number called the dividend is divided by another, called the divisor. The answer to the problem would be the quotient, and in the case of Euclidean division, the remainder would be included as well. Using short division, arbitrarily large dividends can be handled. [1]