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For instance, if the one solving the math word problem has a limited understanding of the language (English, Spanish, etc.) they are more likely to not understand what the problem is even asking. In Example 1 (above), if one does not comprehend the definition of the word "spent," they will misunderstand the entire purpose of the word problem.
A simple fraction (also known as a common fraction or vulgar fraction) [n 1] is a rational number written as a/b or , where a and b are both integers. [9] As with other fractions, the denominator (b) cannot be zero. Examples include 1 / 2 , − 8 / 5 , −8 / 5 , and 8 / −5 .
If the divisor has a fractional part, one can restate the problem by moving the decimal to the right in both numbers until the divisor has no fraction, which can make the problem easier to solve (e.g., 10/2.5 = 100/25 = 4). Division can be calculated with an abacus. [14]
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.
For algorithms describing how to calculate the remainder, see Division algorithm.) The remainder, as defined above, is called the least positive remainder or simply the remainder . [ 2 ] The integer a is either a multiple of d , or lies in the interval between consecutive multiples of d , namely, q ⋅ d and ( q + 1) d (for positive q ).
In computing, the modulo operation returns the remainder or signed remainder of a division, after one number is divided by another, called the modulus of the operation.. Given two positive numbers a and n, a modulo n (often abbreviated as a mod n) is the remainder of the Euclidean division of a by n, where a is the dividend and n is the divisor.
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.
A fraction that is reducible can be reduced by dividing both the numerator and denominator by a common factor. It can be fully reduced to lowest terms if both are divided by their greatest common divisor. [5] In order to find the greatest common divisor, the Euclidean algorithm or prime factorization can be used. The Euclidean algorithm is ...
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