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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.
For example, the expression "5 mod 2" evaluates to 1, because 5 divided by 2 has a quotient of 2 and a remainder of 1, while "9 mod 3" would evaluate to 0, because 9 divided by 3 has a quotient of 3 and a remainder of 0. Although typically performed with a and n both being integers, many computing systems now allow other types of numeric operands.
31.75 4)127.00 12 (12 ÷ 4 = 3) 07 (0 remainder, bring down next figure) 4 (7 ÷ 4 = 1 r 3) 3.0 (bring down 0 and the decimal point) 2.8 (7 × 4 = 28, 30 ÷ 4 = 7 r 2) 20 (an additional zero is brought down) 20 (5 × 4 = 20) 0
43 = (−8) × (−5) + 3, and 3 is the least positive remainder, while, 43 = (−9) × (−5) + (−2) and −2 is the least absolute remainder. In the division of 42 by 5, we have: 42 = 8 × 5 + 2, and since 2 < 5/2, 2 is both the least positive remainder and the least absolute remainder. In these examples, the (negative) least absolute ...
This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
17 is divided into 3 groups of 5, with 2 as leftover. Here, the dividend is 17, the divisor is 3, the quotient is 5, and the remainder is 2 (which is strictly smaller than the divisor 3), or more symbolically, 17 = (3 × 5) + 2.
If one knows that the remainder of n divided by 3 is 2, the remainder of n divided by 5 is 3, and the remainder of n divided by 7 is 2, then with no other information, one can determine the remainder of n divided by 105 (the product of 3, 5, and 7) without knowing the value of n. In this example, the remainder is 23.
Divide the highest term of the remainder by the highest term of the divisor (3x ÷ x = 3). Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (− ...