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 "1971 AHSME Problems/Problem 13"

(created solution page)
 
(→‎Solution)
 
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
−
<math>\boxed{\textbf{(E) }8}</math>.
+
We can rewrite <math>(1.0025)^{10}</math> as <math>(1+\tfrac{25}{10^4})^{10}</math>. By the [[Binomial Theorem]], we know that this expression equals <math>1+ \tbinom{10}1\cdot\tfrac{25}{10^4}+\tbinom{10}2\cdot\tfrac{25^2}{10^8}+\tbinom{10}3\cdot\tfrac{25^3}{10^{12}}+\cdots+(\tfrac{25}{10^4})^{10}</math>. We are looking for the <math>\tfrac1{10^5}</math> term (the fifth decimal digit), so we can disregard the first two terms, because they will not affect our final result. Thus, we are left with <math>\tbinom{10}2\cdot\tfrac{25^2}{10^8}+\tbinom{10}3\cdot\tfrac{25^3}{10^{12}}+\cdots = \tfrac{45\cdot25^2}{10^8}+\tfrac{120\cdot25^3}{10^{12}}+\cdots = \tfrac{28,125}{10^8}+\tfrac{1,875,000}{10^{12}}+\cdots</math>. The first term here has a <math>\tfrac1{10^5}</math> term of <math>8</math>, and the second term does not contribute to the fifth decimal place, because it does not have enough digits in the numerator. So, we can disregard the subsequent terms, because they will not have enough digits either. Thus, the fifth digit of the expansion is <math>\boxed{\textbf{(E) }8}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AHSME 35p box|year=1971|num-b=12|num-a=14}}
 
{{AHSME 35p box|year=1971|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:48, 1 August 2024

Problem

If $(1.0025)^{10}$ is evaluated correct to $5$ decimal places, then the digit in the fifth decimal place is

$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }5\qquad \textbf{(E) }8$

Solution

We can rewrite $(1.0025)^{10}$ as $(1+\tfrac{25}{10^4})^{10}$. By the Binomial Theorem, we know that this expression equals $1+ \tbinom{10}1\cdot\tfrac{25}{10^4}+\tbinom{10}2\cdot\tfrac{25^2}{10^8}+\tbinom{10}3\cdot\tfrac{25^3}{10^{12}}+\cdots+(\tfrac{25}{10^4})^{10}$. We are looking for the $\tfrac1{10^5}$ term (the fifth decimal digit), so we can disregard the first two terms, because they will not affect our final result. Thus, we are left with $\tbinom{10}2\cdot\tfrac{25^2}{10^8}+\tbinom{10}3\cdot\tfrac{25^3}{10^{12}}+\cdots = \tfrac{45\cdot25^2}{10^8}+\tfrac{120\cdot25^3}{10^{12}}+\cdots = \tfrac{28,125}{10^8}+\tfrac{1,875,000}{10^{12}}+\cdots$. The first term here has a $\tfrac1{10^5}$ term of $8$, and the second term does not contribute to the fifth decimal place, because it does not have enough digits in the numerator. So, we can disregard the subsequent terms, because they will not have enough digits either. Thus, the fifth digit of the expansion is $\boxed{\textbf{(E) }8}$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 • 31 • 32 • 33 • 34 • 35
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=1971_AHSME_Problems/Problem_13&oldid=224146"