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Difference between revisions of "1985 AHSME Problems/Problem 30"

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<cmath> 4a^2-32a+55>0 </cmath>
 
<cmath> 4a^2-32a+55>0 </cmath>
  
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The first inequality simplifies to <math>(2a-10)^2\le 49</math>, or <math>|2a-10|\le 7</math>. Since <math>2a-10</math> is even, we must have <math>|2a-10|\le 6</math>, so <math>|a-5|\le 3</math>. Therefore, <math>2\le a\le 8</math>. The second inequality simplifies to <math>(2a-8)^2 > 9</math>, or <math>|2a-8| > 3</math>. Therefore, as <math>2a-8</math> is even, we have <math>|2a-8|\ge 4</math>, or <math>|a-4|\ge 2</math>. Hence <math>a\ge 6</math> or <math>a\le 2</math> Since both inequalities must be satisfied, we see that only <math>a=2</math>, <math>a=6</math>, <math>a=7</math>, and <math>a=8</math> satisfy both inequalities. Each will lead to a distinct solution for <math>n</math>, and thus for <math>x</math>, for a total of <math>4</math> positive solutions.
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The first inequality simplifies to <math>(2a-10)^2\le 49</math>, or <math>|2a-10|\le 7</math>. Since <math>2a-10</math> is even, we must have <math>|2a-10|\le 6</math>, so <math>|a-5|\le 3</math>. Therefore, <math>2\le a\le 8</math>. The second inequality simplifies to <math>(2a-8)^2 > 9</math>, or <math>|2a-8| > 3</math>. Therefore, as <math>2a-8</math> is even, we have <math>|2a-8|\ge 4</math>, or <math>|a-4|\ge 2</math>. Hence <math>a\ge 6</math> or <math>a\le 2</math>. Since both inequalities must be satisfied, we see that only <math>a=2</math>, <math>a=6</math>, <math>a=7</math>, and <math>a=8</math> satisfy both inequalities. Each will lead to a distinct solution for <math>n</math>, and thus for <math>x</math>, for a total of <math>4</math> positive solutions.
  
 
Now let <math> x=-\frac{\sqrt{n}}{2} </math>. We have
 
Now let <math> x=-\frac{\sqrt{n}}{2} </math>. We have

Revision as of 00:22, 3 April 2018

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

See Also

1985 AHSME (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

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