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Let [x] be the greatest integer not exceeding x and {x} the fractional part of x. How many real numbers x satisfy (x - 2)[x] = {x} - 1?

1. 0
2. 1
3. 2
4. infinite

Correct answer: infinite

Solution

Put x = n + f. The equation (x-2)[x] = {x} - 1 becomes (n + f - 2)*n = f - 1. For n = 1: (f - 1) = f - 1, which holds for every f in [0,1), i.e. for all x in [1, 2). That alone gives infinitely many solutions, so the answer is infinite.

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