Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ How many real numbers $x$ satisfy the equation $x^2 + 10,000 \lfloor x \rfloor = 10,000x$ ?

200
199
201
198
197

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let x=a+b where a is the integer part of x and b is the decimal. So the equation will become (a+b)^2+10,000a=10,000a+10,000b. Simplifying we get that (a+b)^2=10,000b, and taking the square root of both sides give us a+b = ±100√b. If a+b = 100√b then a+b must be less than 100, so our maximum value is 99 since we are only looking at integers. If a+b = -100√b then a+b must be greater than -100, giving us a minimum of -99. So the solutions are all integers from -99 to 99 inclusive.