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Difference between revisions of "1997 AJHSME Problems/Problem 6"
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The digit "9" is 6 places to the left of the digit "3". Thus, it has a place value that is <math>10^{6-1}</math> times greater.
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==Problem==
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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?
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<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>
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==Solution 1==
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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>
 
<math>\boxed{\textbf{(C)}}</math>
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==Solution 2==
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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>
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== See also ==
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{{AJHSME box|year=1997|num-b=5|num-a=7}}
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* [[AJHSME]]
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* [[AJHSME Problems and Solutions]]
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* [[Mathematics competition resources]]

Revision as of 13:31, 31 July 2011

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
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