1985 AHSME Problems/Problem 30

Problem

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

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

Solution

We can rearrange the equation into $4x^2=40\lfloor x \rfloor-51$ . Obviously, the RHS is an integer, so $4x^2=n$ for some integer $n$ . We can therefore make the substitution $x=\frac{\sqrt{n}}{2}$ to get

$40\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor-51=n$

(We'll try the case where $x=-\frac{\sqrt{n}}{2}$ later.) Now let $a\le\frac{\sqrt{n}}{2}<a+1$ for a nonnegative integer $a$ , so that $\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor=a$ .

$40a-51=n$

Going back to $a\le\frac{\sqrt{n}}{2}<a+1$ , this implies $4a^2\le n<4a^2+8a+4$ . Making the substitution $n=40a-51$ gives the system of inequalities

$4a^2-40a+51\le0$ $4a^2-32a+55>0$

The first inequality simplifies to $(2a-10)^2\le 49$ , or $|2a-10|\le 7$ . Since $2a-10$ is even, we must have $|2a-10|\le 6$ , so $|a-5|\le 3$ . Therefore, $2\le a\le 8$ . The second inequality simplifies to $(2a-8)^2 > 9$ , or $|2a-8| > 3$ . Therefore, as $2a-8$ is even, we have $|2a-8|\ge 4$ , or $|a-4|\ge 2$ . Hence $a\ge 6$ or $a\le 2$ Since both inequalities must be satisfied, we see that only $a=2$ , $a=6$ , $a=7$ , and $a=8$ satisfy both inequalities. Each will lead to a distinct solution for $n$ , and thus for $x$ , for a total of $4$ positive solutions.

Now let $x=-\frac{\sqrt{n}}{2}$ . We have

$40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51=n$

Since $\lfloor -x\rfloor=-\lceil x\rceil$ , this 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 $1$ , hence $-40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51\le -91$ , i.e., it must be negative. But it is also equal to $n$ , which is positive, and this is a contradiction. Therefore, there are no negative roots.

The total number of roots to this equation is thus $4$ , or $\boxed{(\text{E})}$ .

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

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