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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.
Suppose that a pie has 9 slices and they are to be divided evenly among 4 people. Using Euclidean division, 9 divided by 4 is 2 with remainder 1. In other words, each person receives 2 slices of pie, and there is 1 slice left over. This can be confirmed using multiplication, the inverse of division: if each of the 4 people received 2 slices ...
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.
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.
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.
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 remainder is obtained from the least positive remainder by subtracting 5 ...
The multiplicative inverse x ≡ a −1 (mod m) may be efficiently computed by solving Bézout's equation a x + m y = 1 for x, y, by using the Extended Euclidean algorithm. In particular, if p is a prime number, then a is coprime with p for every a such that 0 < a < p ; thus a multiplicative inverse exists for all a that is not congruent to ...
Place the result (+x) below the bar. x 2 has been divided leaving no remainder, and can therefore be marked as used. The result x is then multiplied by the second term in the divisor −3 = −3x. Determine the partial remainder by subtracting 0x − (−3x) = 3x. Mark 0x as used and place the new remainder 3x above it.