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Suppose a real number P is such that [(1/2) p] + [(1/3) p] + [(1/5) p] = p Here, [x] denot
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Writing the expression as f(p) we note that f(p+30) = 31+f(p).
Evaluating for numbers p lying between 0 and 29 and using the above result you get 28 solutions.
But all this is pretty much brute force. I wonder if there are prettier methods
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