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let f(x) be a polynomial with integral co - efficients such that f(m) = f(n) = f(o) = f(p) = 2008 where m ≠n ≠o ≠p prove that there will exist no integer k such that f(k)=2010 ...
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Solve for x 1/[x] + 1/[2x] = {x} + 1/3 [x] - greatest integer function {x} - fractional part of x ...
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Find the solutions for x and y x + y=[x][y] [.] - stands for greatest integer function ...
