Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(Improved solution, formatting, and LaTeX)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Let <math> \lfloor x \rfloor </math> be the greatest integer less than or equal to <math> x </math>. Then the number of real solutions to <math> 4x^2-40\lfloor x \rfloor +51=0 </math> is
+
Let <math>\left\lfloor x\right\rfloor</math> be the greatest integer less than or equal to <math>x</math>. Then the number of real solutions to <math>4x^2-40\left\lfloor x\right\rfloor+51 = 0</math> is
  
 
<math> \mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 2 \qquad \mathrm{(D) \  } 3 \qquad \mathrm{(E) \  }4 </math>
 
<math> \mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 2 \qquad \mathrm{(D) \  } 3 \qquad \mathrm{(E) \  }4 </math>
  
 
==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 rearrange the equation as <math>4x^2 = 40\left\lfloor x\right\rfloor-51</math>, where the right-hand side is now clearly an integer, meaning that <math>4x^2 = n</math> for some non-negative integer <math>n</math>. Therefore, in the case where <math>x \geq 0</math>, substituting <math>x = \frac{\sqrt{n}}{2}</math> gives
 +
<cmath>40\left\lfloor\frac{\sqrt{n}}{2}\right\rfloor-51 = n.</cmath> To proceed, let <math>a</math> be the unique non-negative integer such that <math>a \leq \frac{\sqrt{n}}{2} < a+1</math>, so that <cmath>\begin{align*}&\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor = a, \text{ and} \\ &4a^2 \leq n < 4a^2+8a+4,\end{align*}</cmath> and our equation reduces to <cmath>40a-51 = n.</cmath>
  
−
<cmath> 40\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor-51=n </cmath>
+
The above inequalities therefore become <cmath>4a^2 \leq 40a-51 < 4a^2+8a+4
 +
\iff 4a^2-40a+51 < 0 \text{ and } 4a^2-32a+55 > 0,</cmath> where the first inequality can now be rewritten as <math>(2a-10)^2 \leq 49</math>, i.e. <math>\left\lvert 2a-10\right\rvert \leq 7</math>. Since <math>(2a-10)</math> is even for all integers <math>a</math>, we must in fact have <cmath>\begin{align*}\left\lvert 2a-10\right\rvert \leq 6 &\iff \left\lvert a-5\right\rvert \leq 3 \\ &\iff 2 \leq a \leq 8.\end{align*}</cmath> The second inequality similarly simplifies to <math>(2a-8)^2 > 9</math>, i.e. <math>\left\lvert 2a-8\right\rvert > 3</math>. As <math>(2a-8)</math> is even, this is equivalent to <cmath>\begin{align*}\left\lvert 2a-8 \right\rvert \geq 4 &\iff \left\lvert a-4\right\rvert \geq 2 \\ &\iff a \geq 6 \text{ or } a \leq 2,\end{align*}</cmath> so the values of <math>a</math> satisfying both inequalities are <math>2</math>, <math>6</math>, <math>7</math>, and <math>8</math>. Since <math>n = 40a-51</math>, each of these distinct values of <math>a</math> gives a distinct solution for <math>n</math>, and thus for <math>x = \frac{\sqrt{n}}{2}</math>, giving a total of <math>4</math> solutions in the <math>x \geq 0</math> case.
  
−
(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 a nonnegative integer <math> a </math>, so that <math> \left\lfloor \frac{\sqrt{n}}{2}\right\rfloor=a </math>.
 
  
−
<cmath> 40a-51=n </cmath>
+
As <math>4</math> is already the largest of the answer choices, this suffices to show that the answer is <math>\text{(E)}</math>, but for completeness, we will show that the <math>x < 0</math> case indeed gives no other solutions. If <math>x = -\frac{\sqrt{n}}{2}</math> (and so <math>n > 0</math>), we require <cmath>40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51 = n,</cmath> and recalling that <math>\left\lfloor -x\right\rfloor = -\left\lceil x\right\rceil</math> for all <math>x</math>, this equation can be rewritten as <cmath>-40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51 = n.</cmath> Since <math>n</math> is positive, the least possible value of <math>\left\lceil \frac{\sqrt{n}}{2}\right\rceil</math> is <math>1</math>, but this means <cmath>\begin{align*}n &= -40\left\lceil\frac{\sqrt{n}}{2}\right\rceil-51 \\ &\leq -40 \cdot 1 - 51 \\ &= -91,\end{align*}</cmath> which is a contradiction. Therefore the <math>x < 0</math> case indeed gives no further solutions, confirming that the total number of solutions is precisely <math>\boxed{\text{(E)} \ 4}</math>.
−
 
 
−
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^2-32a+55>0 </cmath>
 
−
 
 
−
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
 
−
 
 
−
<cmath> 40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51=n </cmath>
 
−
 
 
−
Since <math> \lfloor -x\rfloor=-\lceil x\rceil </math>, this can be rewritten as
 
−
 
 
−
<cmath> -40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51=n </cmath>
 
−
 
 
−
Since <math> n </math> is positive, the least possible value of <math> \left\lceil \frac{\sqrt{n}}{2}\right\rceil </math> is <math> 1 </math>, hence <math> -40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51\le -91</math>, i.e., it must be negative. But it is also equal to <math>n</math>, which is positive, and this is a contradiction. Therefore, there are no negative roots.
 
−
 
 
−
The total number of roots to this equation is thus <math> 4</math>, or <math>\boxed{(\text{E})} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=29|after=Last Problem}}
 
{{AHSME box|year=1985|num-b=29|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:57, 20 March 2024

Problem

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

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

Solution

We rearrange the equation as $4x^2 = 40\left\lfloor x\right\rfloor-51$, where the right-hand side is now clearly an integer, meaning that $4x^2 = n$ for some non-negative integer $n$. Therefore, in the case where $x \geq 0$, substituting $x = \frac{\sqrt{n}}{2}$ gives \[40\left\lfloor\frac{\sqrt{n}}{2}\right\rfloor-51 = n.\] To proceed, let $a$ be the unique non-negative integer such that $a \leq \frac{\sqrt{n}}{2} < a+1$, so that \begin{align*}&\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor = a, \text{ and} \\ &4a^2 \leq n < 4a^2+8a+4,\end{align*} and our equation reduces to \[40a-51 = n.\]

The above inequalities therefore become \[4a^2 \leq 40a-51 < 4a^2+8a+4 \iff 4a^2-40a+51 < 0 \text{ and } 4a^2-32a+55 > 0,\] where the first inequality can now be rewritten as $(2a-10)^2 \leq 49$, i.e. $\left\lvert 2a-10\right\rvert \leq 7$. Since $(2a-10)$ is even for all integers $a$, we must in fact have \begin{align*}\left\lvert 2a-10\right\rvert \leq 6 &\iff \left\lvert a-5\right\rvert \leq 3 \\ &\iff 2 \leq a \leq 8.\end{align*} The second inequality similarly simplifies to $(2a-8)^2 > 9$, i.e. $\left\lvert 2a-8\right\rvert > 3$. As $(2a-8)$ is even, this is equivalent to \begin{align*}\left\lvert 2a-8 \right\rvert \geq 4 &\iff \left\lvert a-4\right\rvert \geq 2 \\ &\iff a \geq 6 \text{ or } a \leq 2,\end{align*} so the values of $a$ satisfying both inequalities are $2$, $6$, $7$, and $8$. Since $n = 40a-51$, each of these distinct values of $a$ gives a distinct solution for $n$, and thus for $x = \frac{\sqrt{n}}{2}$, giving a total of $4$ solutions in the $x \geq 0$ case.


As $4$ is already the largest of the answer choices, this suffices to show that the answer is $\text{(E)}$, but for completeness, we will show that the $x < 0$ case indeed gives no other solutions. If $x = -\frac{\sqrt{n}}{2}$ (and so $n > 0$), we require \[40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51 = n,\] and recalling that $\left\lfloor -x\right\rfloor = -\left\lceil x\right\rceil$ for all $x$, this equation 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 $1$, but this means \begin{align*}n &= -40\left\lceil\frac{\sqrt{n}}{2}\right\rceil-51 \\ &\leq -40 \cdot 1 - 51 \\ &= -91,\end{align*} which is a contradiction. Therefore the $x < 0$ case indeed gives no further solutions, confirming that the total number of solutions is precisely $\boxed{\text{(E)} \ 4}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_30&oldid=217187"