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Answer: 7 × 1 + 6 × 10 + 5 × 9 + 4 × 12 + 3 × 3 + 2 × 4 + 1 × 1 = 178 mod 13 = 9 Remainder = 9 A recursive method can be derived using the fact that = and that =. This implies that a number is divisible by 13 iff removing the first digit and subtracting 3 times that digit from the new first digit yields a number divisible by 13.
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 the quotient.
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 this case, s is called the least absolute remainder. [3] As with the quotient and remainder, k and s are uniquely determined, except in the case where d = 2n and s = ±n. For this exception, we have: a = kd + n = (k + 1)d − n. A unique remainder can be obtained in this case by some convention—such as always taking the positive value of s.
Here, the partial dividend is 9. The first number to be divided by the divisor (4) is the partial dividend (9). One writes the integer part of the result (2) above the division bar over the leftmost digit of the dividend, and one writes the remainder (1) as a small digit above and to the right of the partial dividend (9).
127 ÷ 4 = 31.75 124 30 (bring down 0; decimal to quotient) 28 (7 × 4 = 28) 20 (an additional zero is added) 20 (5 × 4 = 20) 0 In Mexico , the English-speaking world notation is used, except that only the result of the subtraction is annotated and the calculation is done mentally, as shown below:
7 + 9 → 6, carry 1 (since 7 + 9 = 16 = 6 + (1 × 10 1)) This is known as carrying . [ 41 ] When the result of an addition exceeds the value of a digit, the procedure is to "carry" the excess amount divided by the radix (that is, 10/10) to the left, adding it to the next positional value.
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 − (− ...