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1985 AHSME Problems/Problem 30

Revision as of 22:04, 14 February 2014 by Sanqiang3 (talk | contribs) (Switched +/- based on problem statement)

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

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

Approximating the roots of these two quadratics gives two integral solutions for $a$. 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 $2, \boxed{\text{C}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
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