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 "1997 AJHSME Problems/Problem 6"

(Created page with "The digit "9" is 6 places to the left of the digit "3". Thus, it has a place value that is <math>10^6</math> times greater. <math>\boxed{\textbf{(D)}}</math>")
 
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
−
The digit "9" is 6 places to the left of the digit "3". Thus, it has a place value that is <math>10^6</math> times greater.
+
==Problem==
−
<math>\boxed{\textbf{(D)}}</math>
+
 
 +
In the number <math>74982.1035</math> the value of the ''place'' occupied by the digit 9 is how many times as great as the value of the ''place'' occupied by the digit 3?
 +
 
 +
<math>\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000</math>
 +
 
 +
==Solution 1==
 +
The digit <math>9</math> is <math>5</math> places to the left of the digit <math>3</math>. Thus, it has a place value that is <math>10^5 = 100,000</math> times greater.
 +
<math>\boxed{\textbf{(C)}}</math>
 +
 
 +
==Solution 2==
 +
The digit <math>9</math> is in the <math>100</math>s place.  The digit <math>3</math> is in the <math>\frac{1}{1000}</math>ths place.  The ratio of these two numbers is <math>\frac{100}{\frac{1}{1000}} = 100\times1000 = 100,000</math>, and the answer is <math>\boxed{\textbf{(C)}}</math>
 +
 
 +
 
 +
== See also ==
 +
{{AJHSME box|year=1997|num-b=5|num-a=7}}
 +
* [[AJHSME]]
 +
* [[AJHSME Problems and Solutions]]
 +
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 23:26, 4 July 2013

Problem

In the number $74982.1035$ the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?

$\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000$

Solution 1

The digit $9$ is $5$ places to the left of the digit $3$. Thus, it has a place value that is $10^5 = 100,000$ times greater. $\boxed{\textbf{(C)}}$

Solution 2

The digit $9$ is in the $100$s place. The digit $3$ is in the $\frac{1}{1000}$ths place. The ratio of these two numbers is $\frac{100}{\frac{1}{1000}} = 100\times1000 = 100,000$, and the answer is $\boxed{\textbf{(C)}}$


See also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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=1997_AJHSME_Problems/Problem_6&oldid=55741"