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 1

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 $\boxed{\text{(E)} \ 4}$ .

Solution 2

This equation is equivalent to $4x^2 - 40x + 40x - 40\lfloor x \rfloor + 51 = 0$ , or $4x^2 - 40x + 51 + 40(x - \lfloor x \rfloor) = 0$ . Let $\{x\} = x - \lfloor x \rfloor$ be the fractional part of $x$ . Then this equation becomes $4x^2 - 40x + 51 + 40\{x\} = 0$ .

Note that for all $x$ , $0 \leq \{x\} < 1$ . Therefore, $4x^2 - 40x + 51 \leq 4x^2 - 40x + 51 + 40\{x\} \leq 4x^2 - 40x + 91$ , that is $4x^2 - 40x + 51 \leq 0 < 4x^2 - 40x + 91$ .

First, since $4x^2 - 40x + 51 = (2x - 17)(2x - 3) \leq 0$ , it follows that $\frac{3}{2} \leq x \leq \frac{17}{2}$ .

Also, since $4x^2 - 40x + 91 = (2x - 13)(2x - 7) > 0$ , either $x < \frac{7}{2}$ or $x > \frac{13}{2}$ .

Thus either $\frac{3}{2} \leq x < \frac{7}{2}$ or $\frac{13}{2} < x \leq \frac{17}{2}$ , so $\lfloor x \rfloor \in \{1, 2, 3, 6, 7, 8\}$ .

Isolating $x^2$ in the original equation yields $x^2 = \frac{40\lfloor x \rfloor - 51}{4}$ .

If $\lfloor x \rfloor = 1$ , then $x^2 = -\frac{11}{4}$ . This does not yield any valid solutions as $x^2$ may not be negative.

If $\lfloor x \rfloor = 2$ , then $x^2 = \frac{29}{4}$ and $x = \frac{\sqrt{29}}{2}$ . This is a valid solution since $\left\lfloor\frac{\sqrt{29}}{2}\right\rfloor = 2$ .

If $\lfloor x \rfloor = 3$ , then $x^2 = \frac{69}{4}$ and $x = \frac{\sqrt{69}}{2}$ . This is not a valid solution since $\left\lfloor\frac{\sqrt{69}}{2}\right\rfloor = 4$ .

If $\lfloor x \rfloor = 6$ , then $x^2 = \frac{189}{4}$ and $x = \frac{\sqrt{189}}{2}$ . This is a valid solution since $\left\lfloor\frac{\sqrt{189}}{2}\right\rfloor = 6$ .

If $\lfloor x \rfloor = 7$ , then $x^2 = \frac{229}{4}$ and $x = \frac{\sqrt{229}}{2}$ . This is a valid solution since $\left\lfloor\frac{\sqrt{229}}{2}\right\rfloor = 7$ .

If $\lfloor x \rfloor = 8$ , then $x^2 = \frac{269}{4}$ and $x = \frac{\sqrt{269}}{2}$ . This is a valid solution since $\left\lfloor\frac{\sqrt{269}}{2}\right\rfloor = 8$ .

Therefore, this equation as a total of $\boxed{(\textbf{E})\ 4}$ solutions.

-j314andrews

1985 AHSME (Problems • Answer Key • Resources)
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1985 AHSME Problems/Problem 30

Contents

Problem

Solution 1

Solution 2

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