1997 AJHSME Problems/Problem 6

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 $\boxed{\textbf{(C)}}$

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1997 AJHSME Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See also