2018 AMC 10B Problem 25

Rewrite the equation as
$$x^2 = 10000\,(x - \lfloor x \rfloor) = 10000\,\{x\},$$
where $\{x\} \in [0, 1)$ is the fractional part. So $0 \le x^2 < 10000$, forcing $-100 < x < 100$.

Write $x = n + f$ with $n = \lfloor x \rfloor$ an integer and $f \in [0, 1)$. For a fixed $n$ we need a root $f \in [0,1)$ of
$$q(f) = (n + f)^2 - 10000 f = f^2 + (2n - 10000)f + n^2.$$
Since $|n| \le 100$, the two roots of $q$ have sum $10000 - 2n > 1$ and product $n^2 \ge 0$, so at most one root can lie in $[0, 1)$; if one root is below $1$ the other exceeds $10000 - 2n - 1 > 1$.

Now $q(0) = n^2 \ge 0$ and $q(1) = (n+1)^2 - 10000$. When $q(1) < 0$, i.e. $|n + 1| < 100$, there is a sign change on $[0, 1)$ and hence exactly one solution. This gives $n = -100, -99, \dots, 98$, which is $199$ integers (for $n = 0$ the solution is $f = 0$, i.e. $x = 0$, which does satisfy the equation).

Boundary cases: $n = 99$ gives $q(f) = (f - 1)(f - 9801)$, whose only small root is $f = 1$, not allowed. $n = -101$ would force $x < -100$, excluded. Larger $|n|$ is impossible.

Total: $199$ solutions. The answer is $\boxed{\textbf{(C)}\ 199}$.

Floor equations become ordinary equations on each interval $[n, n+1)$ once you write $x = n + f$. Then the question is how many of these intervals contain a root, which a sign check at the endpoints answers in one stroke. The whole difficulty is at the edges, where the root lands exactly on $f = 1$ and is lost; check those cases separately.

Graphically, the equation is $\{x\} = \dfrac{x^2}{10000}$. The right side is a gentle parabola with values in $[0, 1)$ for $|x| < 100$; the left side is the sawtooth rising from $0$ to $1$ on each $[n, n+1)$. Each tooth crosses the parabola once when the parabola is below $1$ at the tooth's right end, i.e. $(n+1)^2 < 10000$: the teeth $n = -100, \dots, 98$, which is $199$ crossings. On $[99, 100)$ the parabola stays above the tooth.

• Counting one solution for every integer n with |x| < 100 (n = -100 to 99, giving 200) without noticing that n = 99 yields only f = 1, which is excluded.
• Dropping negative $x$, or dropping $x = 0$ (where $f = 0$ is legitimate), which gives $99$ or $198$.