# 1985 AHSME Problems/Problem 30

## Problem

Let $\left\lfloor x\right\rfloor$ be the greatest integer less than or equal to $x$. Then the number of real solutions to $4x^2-40\left\lfloor x\right\rfloor+51 = 0$ is

$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4$

## Solution

We rearrange the equation as $4x^2 = 40\left\lfloor x\right\rfloor-51$, where the right-hand side is now clearly an integer, meaning that $4x^2 = n$ for some non-negative integer $n$. Therefore, in the case where $x \geq 0$, substituting $x = \frac{\sqrt{n}}{2}$ gives $$40\left\lfloor\frac{\sqrt{n}}{2}\right\rfloor-51 = n.$$ To proceed, let $a$ be the unique non-negative integer such that $a \leq \frac{\sqrt{n}}{2} < a+1$, so that \begin{align*}&\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor = a, \text{ and} \\ &4a^2 \leq n < 4a^2+8a+4,\end{align*} and our equation reduces to $$40a-51 = n.$$

The above inequalities therefore become $$4a^2 \leq 40a-51 < 4a^2+8a+4 \iff 4a^2-40a+51 < 0 \text{ and } 4a^2-32a+55 > 0,$$ where the first inequality can now be rewritten as $(2a-10)^2 \leq 49$, i.e. $\left\lvert 2a-10\right\rvert \leq 7$. Since $(2a-10)$ is even for all integers $a$, we must in fact have \begin{align*}\left\lvert 2a-10\right\rvert \leq 6 &\iff \left\lvert a-5\right\rvert \leq 3 \\ &\iff 2 \leq a \leq 8.\end{align*} The second inequality similarly simplifies to $(2a-8)^2 > 9$, i.e. $\left\lvert 2a-8\right\rvert > 3$. As $(2a-8)$ is even, this is equivalent to \begin{align*}\left\lvert 2a-8 \right\rvert \geq 4 &\iff \left\lvert a-4\right\rvert \geq 2 \\ &\iff a \geq 6 \text{ or } a \leq 2,\end{align*} so the values of $a$ satisfying both inequalities are $2$, $6$, $7$, and $8$. Since $n = 40a-51$, each of these distinct values of $a$ gives a distinct solution for $n$, and thus for $x = \frac{\sqrt{n}}{2}$, giving a total of $4$ solutions in the $x \geq 0$ case.

As $4$ is already the largest of the answer choices, this suffices to show that the answer is $\text{(E)}$, but for completeness, we will show that the $x < 0$ case indeed gives no other solutions. If $x = -\frac{\sqrt{n}}{2}$ (and so $n > 0$), we require $$40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51 = n,$$ and recalling that $\left\lfloor -x\right\rfloor = -\left\lceil x\right\rceil$ for all $x$, this equation can be rewritten as $$-40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51 = n.$$ Since $n$ is positive, the least possible value of $\left\lceil \frac{\sqrt{n}}{2}\right\rceil$ is $1$, but this means \begin{align*}n &= -40\left\lceil\frac{\sqrt{n}}{2}\right\rceil-51 \\ &\leq -40 \cdot 1 - 51 \\ &= -91,\end{align*} which is a contradiction. Therefore the $x < 0$ case indeed gives no further solutions, confirming that the total number of solutions is precisely $\boxed{\text{(E)} \ 4}$.