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

$40\lfloor \frac{\sqrt{n}}{2}\rfloor+51=n$

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

$40a+51=n$

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

$4a^2-40a-51\le0$ $4a^2-32a-47>0$

Approximating the roots of these two quadratics gives two integral solutions for $a$ : $a=10, 11$ . Each of these gives a distinct solution for $n$ , and thus $x$ , for a total of $2$ positive solutions.

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

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

Since $\lfloor -x\rfloor=-\lceil x\rceil$ , this can be rewritten as

$-40\lceil \frac{\sqrt{n}}{2}\rceil+51=n$

Since $n$ is positive, the only possible value of $\lceil \frac{\sqrt{n}}{2}\rceil$ is $1$ , meaning $n=11$ . However, this would make $\lceil \frac{\sqrt{n}}{2}\rceil=2$ , a contradiction. Therefore, there are no negative roots.

The total number of roots to this equation is thus $2, \boxed{\text{C}}$ .

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

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