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To calculate the whole number quotient of dividing a large number by a small number, the student repeatedly takes away "chunks" of the large number, where each "chunk" is an easy multiple (for example 100×, 10×, 5× 2×, etc.) of the small number, until the large number has been reduced to zero – or the remainder is less than the small ...
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]
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 ...
It breaks down a division problem into a series of easier steps. As in all division problems, one number, called the dividend, is divided by another, called the divisor, producing a result called the quotient. It enables computations involving arbitrarily large numbers to be performed by following a series of simple steps. [1]
To calculate a Pythagorean triple, take any term of this sequence and convert it to an improper fraction (for mixed number , the corresponding improper fraction is ). Then its numerator and denominator are the sides, b and a , of a right triangle, and the hypotenuse is b + 1 .
This procedure solves the fair division problem for two people. The modern study of fair cake-cutting was initiated during World War II, when Hugo Steinhaus asked his students Stefan Banach and Bronisław Knaster to find a generalization of divide-and-choose to three or more people.
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.
An item whose delay is times the length of a message must occupy a fraction of at least / of the time slots on the channel it is assigned to, so a solution to the scheduling problem can only come from a solution to the unit fraction bin packing problem with the channels as bins and the fractions / as item sizes.
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