No. It has a discontinuity at every integer value.
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There is no greatest integer. Whatever integer you think is greatest, you can always add one (1) to it and get a larger one.
A floor function takes any decimal and rounds it down to the closest integer.(You might possibly also call this a "ground floor" function.)
Solving for the third integer:(a + b + c) / 3 = 37 (by the definition of "average")(18 + b + c) / 3 = 37 (assuming "a" is the least integer)18 + b + c = 111c = 111 - 18 - b (solving for the third integer)c = 93 - bNow, "b" must be at least 18, in which case:c = 93 - 18 = 75On the other hand, the largest "b" can be is when it is equal to "c" (since I am assuming that "c" is the largest integer):c = 93 - bc = 93 - c2c = 93c = 46.5Adjusting the numbers a bit (since we need integers), in this case we get the numbers 18, 46, 47Thus, the greatest integer can be anything in the range from 47 to 75.
An algebraic function is a function built from polynomial and combined with +,*,-,/ signs. The transcendental it is not built from polynomial like X the power of Pie plus 1. this function is transcendental because the power pi is not integer number in result it can't be a polynomial.
Let me first re-phrase your question: What is the number of (positive) integers less than 10000 (5 digits) and greater than 999 (3 digits)? The greatest 4 digit integer would be 9999. The greatest 3 digit integer would be 999. Let's do some subtraction: 9999 - 999 = 9000 This works because as we count up from 999, each positive integer encountered satisfies your requirements until reaching 10000.