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At the same time the student is generating a list of the multiples of the small number (i.e., partial quotients) that have so far been taken away, which when added up together would then become the whole number quotient itself. For example, to calculate 132 ÷ 8, one might successively subtract 80, 40 and 8 to leave 4: 132 80 (10 × 8) -- 52 40 ...
If there is a remainder in solving a partition problem, the parts will end up with unequal sizes. For example, if 52 cards are dealt out to 5 players, then 3 of the players will receive 10 cards each, and 2 of the players will receive 11 cards each, since = +.
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.
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).
Such a quadratic irrational may also be written in another form with a square-root of a square-free number (for example (+) /) as explained for quadratic irrationals. By considering the complete quotients of periodic continued fractions, Euler was able to prove that if x is a regular periodic continued fraction, then x is a quadratic irrational ...
Zaremba has conjectured the existence of an absolute constant A, such that the rationals with partial quotients restricted by A contain at least one for every (positive integer) denominator. The choice A = 5 is compatible with the numerical evidence. [4] Further conjectures reduce that value, in the case of all sufficiently large denominators. [5]
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In cases where one or more of the b terms has more than two digits, the final quotient value b cannot be constructed simply by concatenating the digit pairs. Instead, each term, starting with b 1 , {\displaystyle b_{1},} should be multiplied by 100, and the next term added (or, if negative, subtracted).
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