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 "1986 AJHSME Problems/Problem 7"

m (→‎Solution)
(Linear Approximation using Calculus)
 
(2 intermediate revisions by 2 users not shown)
Line 21: Line 21:
  
 
<math>\boxed{\text{B}}</math>
 
<math>\boxed{\text{B}}</math>
 +
 +
==Linear Approximation using Calculus==
 +
While this may not be the best, most efficient way to solve this problem, it can prove to be quite useful in similar problems. Using Calculus, we can use tangent-line approximation to approximate these square roots. First, we understand that these roots are of the form <math>f(x) = \sqrt{x}.</math> Given the values to approximate, 8, and 80, we find the closest square number to them, namely, 9, and 81.
 +
 +
Now, we take the derivative of <math>f(x)</math> to get <math>f'(x) = \dfrac{1}{2\sqrt{x}}.</math> We can approximate <math>f(x)</math>, to get <math>f(x) \approxeq f(x) + f`(x) \cdot \Delta{x},</math> where <math>x</math> is the perfect square closest to the the input value (in this case 8), and <math>\Delta{x}</math> is the difference between <math>x</math> and the input value.
 +
 +
The two cases are as follows:
 +
 +
 +
\begin{array}{|c|c|c|}
 +
input & x\text{ }value & \Delta{x} \\
 +
\hline
 +
8 & 9 & 8-9=-1 \\
 +
\hline
 +
80 & 81 & 80-81=-1 \\
 +
\end{array}
 +
 +
Now, we can plug in our values into the equation, getting:
 +
 +
 +
\begin{align*}
 +
&f(8) \approxeq f(9) + f`(9) \cdot -1 \implies \sqrt{9} + \dfrac{1}{2\sqrt{9}} \cdot -1  \implies 3 - \dfrac{1}{6} \\
 +
&f(80) \approxeq f(81) + f`(81) \cdot -1 \implies\sqrt{81} + \dfrac{1}{2\sqrt{81}} \cdot -1  \implies 9 - \dfrac{1}{18} \\
 +
\end{align*}
 +
 +
Thus, we see that our answer lies in the amount of whole numbers between <math>3 - \dfrac{1}{6},</math> and <math>9 - \dfrac{1}{18},</math> and as we are counting whole numbers, we can round each expression to narrow our interval to <math>[3,8].</math> Therefore, we see that our answer is <math>\boxed{6\text{ }(B)}.</math>
 +
 +
Note: this solution is definitely not the fastest solution, and in a competition, you should probably use intuition to see the closest numbers in the interval of those square roots, and solve the problem from there.
 +
~Cheetahboy93
  
 
==See Also==
 
==See Also==
  
−
[[1986 AJHSME Problems]]
+
{{AJHSME box|year=1986|num-b=6|num-a=8}}
 +
[[Category:Introductory Algebra Problems]]
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:29, 11 August 2024

Problem

How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$?

$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution

No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.

Clearly it must be true that for any positive integers $a$, $b$, and $c$ with $a>b>c$, \[\sqrt{a}>\sqrt{b}>\sqrt{c}\]

If we let $a=9$, $b=8$, and $c=4$, then we get \[\sqrt{9}>\sqrt{8}>\sqrt{4}\] \[3>\sqrt{8}>2\]

Therefore, the smallest whole number between $\sqrt{8}$ and $\sqrt{80}$ is $3$.

Similarly, if we let $a=81$, $b=80$, and $c=64$, we get \[\sqrt{81}>\sqrt{80}>\sqrt{64}\] \[9>\sqrt{80}>8\]

So $8$ is the largest whole number between $\sqrt{8}$ and $\sqrt{80}$.

So we know that we just have to find the number of integers from 3 to 8 inclusive. If we subtract 2 from every number in this set (which doesn't change the number of integers in the set at all), we find that now all we need to do is find the number of integers there are from 1 to 6, which is obviously 6.

$\boxed{\text{B}}$

Linear Approximation using Calculus

While this may not be the best, most efficient way to solve this problem, it can prove to be quite useful in similar problems. Using Calculus, we can use tangent-line approximation to approximate these square roots. First, we understand that these roots are of the form $f(x) = \sqrt{x}.$ Given the values to approximate, 8, and 80, we find the closest square number to them, namely, 9, and 81.

Now, we take the derivative of $f(x)$ to get $f'(x) = \dfrac{1}{2\sqrt{x}}.$ We can approximate $f(x)$, to get $f(x) \approxeq f(x) + f`(x) \cdot \Delta{x},$ where $x$ is the perfect square closest to the the input value (in this case 8), and $\Delta{x}$ is the difference between $x$ and the input value.

The two cases are as follows:


\begin{array}{|c|c|c|} input & x\text{ }value & \Delta{x} \\ \hline 8 & 9 & 8-9=-1 \\ \hline 80 & 81 & 80-81=-1 \\ \end{array}

Now, we can plug in our values into the equation, getting:


\begin{align*} &f(8) \approxeq f(9) + f`(9) \cdot -1 \implies \sqrt{9} + \dfrac{1}{2\sqrt{9}} \cdot -1 \implies 3 - \dfrac{1}{6} \\ &f(80) \approxeq f(81) + f`(81) \cdot -1 \implies\sqrt{81} + \dfrac{1}{2\sqrt{81}} \cdot -1 \implies 9 - \dfrac{1}{18} \\ \end{align*}

Thus, we see that our answer lies in the amount of whole numbers between $3 - \dfrac{1}{6},$ and $9 - \dfrac{1}{18},$ and as we are counting whole numbers, we can round each expression to narrow our interval to $[3,8].$ Therefore, we see that our answer is $\boxed{6\text{ }(B)}.$

Note: this solution is definitely not the fastest solution, and in a competition, you should probably use intuition to see the closest numbers in the interval of those square roots, and solve the problem from there. ~Cheetahboy93

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
All AJHSME/AMC 8 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=1986_AJHSME_Problems/Problem_7&oldid=224931"