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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.
Therefore 1+1=T, and T=2. As column 1 had been calculated in the last step, it is known that there isn't a carry in column 2. But, it is also known that there is a carry in column 3 in the first step. Therefore, 2+G≥10. If G is equal to 9, U would equal 1, but this is impossible as O also equals 1. So only G=8 is possible and with 2+8=10+U, U=0.
The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras. [1]
Balinese, for example, currently has a decimal system, with words for 10, 100, and 1000, but has additional simplex numerals for 25 (with a second word for 25 only found in a compound for 75), 35, 45, 50, 150, 175, 200 (with a second found in a compound for 1200), 400, 900, and 1600.
For functions in certain classes, the problem of determining: whether two functions are equal, known as the zero-equivalence problem (see Richardson's theorem); [5] the zeroes of a function; whether the indefinite integral of a function is also in the class. [6] Of course, some subclasses of these problems are decidable.
For example, 6 + 7 = 13 can be derived from 5 + 7 = 12 by adding one more. [36] Making ten: An advanced strategy uses 10 as an intermediate for sums involving 8 or 9; for example, 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14. [36] As students grow older, they commit more facts to memory, and learn to derive other facts rapidly and fluently.
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