2018 AMC 12B Problem #24

Algebra Level 3

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


This problem is part of the 2018 AMC 12B .
200 199 201 198 197

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1 solution

Loillipop Cuber
Apr 16, 2018

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.

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