2018 AMC 10B Problem 25

Move the terms around so the floor sits next to $x$:
$$x^2 = 10000\,\big(x - \lfloor x \rfloor\big).$$
The quantity $x - \lfloor x \rfloor$ is the part of $x$ after its whole part, the "decimal part." For example, for $x = 3.7$ it is $0.7$, and for $x = -2.3$ it is $0.7$ too (since $\lfloor -2.3 \rfloor = -3$). It is always at least $0$ and less than $1$.

1. Limit the size of $x$. The decimal part is less than $1$, so $x^2 < 10000$. That means $x$ is strictly between $-100$ and $100$.
2. Split $x$. Write $x = n + f$, where $n = \lfloor x \rfloor$ is a whole number and $f$ is the decimal part, $0 \le f < 1$. The equation becomes
$$(n + f)^2 = 10000 f.$$
For each whole number $n$, ask how many $f$ from $0$ up to (not including) $1$ solve this.
3. Compare the two sides as $f$ goes from $0$ to $1$. Let the "left" be $(n+f)^2$ and the "right" be $10000f$.
- At $f = 0$: left $= n^2$, right $= 0$. The left is at least as big.
- At $f = 1$: left $= (n+1)^2$, right $= 10000$.

If $(n+1)^2 < 10000$, the right side has overtaken the left by $f = 1$, so they are equal somewhere in between: one solution. (The right side climbs by $10000$ as $f$ goes from $0$ to $1$, far faster than the left side can change when $|n|$ is at most $100$, so they cross only once.)
4. Which $n$ work? $(n+1)^2 < 10000$ means $n + 1$ is between $-100$ and $100$, not including the ends. So $n = -100, -99, \ldots, 98$. That is
$$98 - (-100) + 1 = 199 \text{ values}.$$
(For $n = 0$ the solution is $f = 0$, that is $x = 0$, which does satisfy the equation.)
5. The edge case $n = 99$. Now $(n+1)^2 = 10000$ exactly, so the two sides only meet at $f = 1$, meaning $x = 100$. But $f$ must be less than $1$, and $x = 100$ was already ruled out in step 1. So no solution here.
6. No other $n$. Step 1 says $x$ is between $-100$ and $100$, so $n$ cannot be $-101$ or less, or $100$ or more.

Total: $199$ solutions.

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

Equations with a floor become ordinary equations once you fix the whole part $n$ and let only the decimal part vary. Then you just ask, interval by interval, whether a solution lands inside. Comparing the two sides at the ends of each interval answers that quickly. The real danger is at the edge, where the solution lands exactly on the boundary and does not count.

Think of graphs. The equation says (decimal part of $x$) $= \dfrac{x^2}{10000}$. The left side is a sawtooth: on each interval from $n$ to $n+1$ it climbs from $0$ almost to $1$, then drops. The right side is a very flat U-shaped curve, below $1$ for $x$ between $-100$ and $100$. Each tooth crosses the curve once as long as the curve is still below $1$ at the tooth's right end, meaning $(n+1)^2 < 10000$. That is the teeth $n = -100$ through $98$, giving $199$ crossings. On the tooth from $99$ to $100$ the curve stays above the tooth.

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 $\boxed{\textbf{(C)}\ 199}$.