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

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==Solution==
 
==Solution==
We can rearrange the equation into <math> 4x^2=40\lfloor x \rfloor+51 </math>. Obviously, the RHS is an integer, so <math> 4x^2=n </math> for some integer <math> n </math>. We can therefore make the substitution <math> x=\frac{\sqrt{n}}{2} </math> to get
+
We can rearrange the equation into <math> 4x^2=40\lfloor x \rfloor-51 </math>. Obviously, the RHS is an integer, so <math> 4x^2=n </math> for some integer <math> n </math>. We can therefore make the substitution <math> x=\frac{\sqrt{n}}{2} </math> to get
  
<cmath> 40\lfloor \frac{\sqrt{n}}{2}\rfloor+51=n </cmath>
+
<cmath> 40\lfloor \frac{\sqrt{n}}{2}\rfloor-51=n </cmath>
  
 
(We'll try the case where <math> x=-\frac{\sqrt{n}}{2} </math> later.) Now let <math> a\le\frac{\sqrt{n}}{2}<a+1 </math> for an integer <math> a </math>, so that <math> \lfloor \frac{\sqrt{n}}{2}\rfloor=a </math>.
 
(We'll try the case where <math> x=-\frac{\sqrt{n}}{2} </math> later.) Now let <math> a\le\frac{\sqrt{n}}{2}<a+1 </math> for an integer <math> a </math>, so that <math> \lfloor \frac{\sqrt{n}}{2}\rfloor=a </math>.
  
<cmath> 40a+51=n </cmath>
+
<cmath> 40a-51=n </cmath>
  
Going back to <math> a\le\frac{\sqrt{n}}{2}<a+1 </math>, this implies <math> 4a^2\le n<4a^2+8a+4 </math>. Making the substitution <math> n=40a+51 </math> gives the system of inequalities
+
Going back to <math> a\le\frac{\sqrt{n}}{2}<a+1 </math>, this implies <math> 4a^2\le n<4a^2+8a+4 </math>. Making the substitution <math> n=40a-51 </math> gives the system of inequalities
  
<cmath> 4a^2-40a-51\le0 </cmath>
+
<cmath> 4a^2-40a+51\le0 </cmath>
<cmath> 4a^2-32a-47>0 </cmath>
+
<cmath> 4a^2-32a+55>0 </cmath>
  
Approximating the roots of these two quadratics gives two integral solutions for <math> a </math>: <math> a=10, 11 </math>. Each of these gives a distinct solution for <math> n </math>, and thus <math> x </math>, for a total of <math> 2 </math> positive solutions.
+
Approximating the roots of these two quadratics gives two integral solutions for <math> a </math>. Each of these gives a distinct solution for <math> n </math>, and thus <math> x </math>, for a total of <math> 2 </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
  
<cmath> 40\lfloor -\frac{\sqrt{n}}{2}\rfloor+51=n </cmath>
+
<cmath> 40\lfloor -\frac{\sqrt{n}}{2}\rfloor-51=n </cmath>
  
 
Since <math> \lfloor -x\rfloor=-\lceil x\rceil </math>, this can be rewritten as
 
Since <math> \lfloor -x\rfloor=-\lceil x\rceil </math>, this can be rewritten as
  
<cmath> -40\lceil \frac{\sqrt{n}}{2}\rceil+51=n </cmath>
+
<cmath> -40\lceil \frac{\sqrt{n}}{2}\rceil-51=n </cmath>
  
 
Since <math> n </math> is positive, the only possible value of <math> \lceil \frac{\sqrt{n}}{2}\rceil </math> is <math> 1 </math>, meaning <math> n=11 </math>. However, this would make <math> \lceil \frac{\sqrt{n}}{2}\rceil=2 </math>, a contradiction. Therefore, there are no negative roots.
 
Since <math> n </math> is positive, the only possible value of <math> \lceil \frac{\sqrt{n}}{2}\rceil </math> is <math> 1 </math>, meaning <math> n=11 </math>. However, this would make <math> \lceil \frac{\sqrt{n}}{2}\rceil=2 </math>, a contradiction. Therefore, there are no negative roots.

Revision as of 23:04, 14 February 2014

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 (ProblemsAnswer KeyResources)
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Problem 29
Followed by
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