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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.
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 terms of partition, 20 / 5 means the size of each of 5 parts into which a set of size 20 is divided. For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". 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 ...
Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...
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,
divide a 16-digit number by an 8-digit divisor. Addition or subtraction is performed in a single step, with a turn of the crank. Multiplication and division are performed digit by digit on the multiplier or divisor digits, in a procedure equivalent to the familiar long multiplication and long division procedures taught in school.
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]
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 − (− ...