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The black numbers are the addends, the green number is the carry, and the blue number is the sum. In the rightmost digit, the addition of 9 and 7 is 16, carrying 1 into the next pair of the digit to the left, making its addition 1 + 5 + 2 = 8. Therefore, 59 + 27 = 86.
Find the shortest sequence of digits starting from the left end of the dividend, 500, that the divisor 4 goes into at least once. In this case, this is simply the first digit, 5. The largest number that the divisor 4 can be multiplied by without exceeding 5 is 1, so the digit 1 is put above the 5 to start constructing the quotient.
For example, there are six divisors of 4; they are 1, 2, 4, −1, −2, and −4, but only the positive ones (1, 2, and 4) would usually be mentioned. 1 and −1 divide (are divisors of) every integer. Every integer (and its negation) is a divisor of itself. Integers divisible by 2 are called even, and integers not divisible by 2 are called odd.
In abstract algebra, given a magma with binary operation ∗ (which could nominally be termed multiplication), left division of b by a (written a \ b) is typically defined as the solution x to the equation a ∗ x = b, if this exists and is unique. Similarly, right division of b by a (written b / a) is the solution y to the equation y ∗ a = b ...
Instead, the division is reduced to small steps. Starting from the left, enough digits are selected to form a number (called the partial dividend) that is at least 4×1 but smaller than 4×10 (4 being the divisor in this problem). Here, the partial dividend is 9. The first number to be divided by the divisor (4) is the partial dividend (9).
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