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Note: The reason why this works is that if we have: a+b=c and b is a multiple of any given number n, then a and c will necessarily produce the same remainder when divided by n. In other words, in 2 + 7 = 9, 7 is divisible by 7. So 2 and 9 must have the same remainder when divided by 7. The remainder is 2.
Thus, the function may be more "cheaply" evaluated using synthetic division and the polynomial remainder theorem. The factor theorem is another application of the remainder theorem: if the remainder is zero, then the linear divisor is a factor. Repeated application of the factor theorem may be used to factorize the polynomial. [3]
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.
Suppose that a pie has 9 slices and they are to be divided evenly among 4 people. Using Euclidean division, 9 divided by 4 is 2 with remainder 1. In other words, each person receives 2 slices of pie, and there is 1 slice left over. This can be confirmed using multiplication, the inverse of division: if each of the 4 people received 2 slices ...
In mathematics, the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1).
The rings for which such a theorem exists are called Euclidean domains, but in this generality, uniqueness of the quotient and remainder is not guaranteed. [8] Polynomial division leads to a result known as the polynomial remainder theorem: If a polynomial f(x) is divided by x − k, the remainder is the constant r = f(k). [9] [10]
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 − (− ...
(mod 5) (–4 is the congruence class) So if we began in modulo class –4 nuts then we will remain in modulo class –4. Since ultimately we have to divide the pile 5 times or 5^5, the original pile was 5^5 – 4 = 3121 coconuts. The remainder of 1020 coconuts conveniently divides evenly by 5 in the morning.