2008 AMC 8 Problems/Problem 22

Problem

For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$

Solution

If $\frac{n}{3}$ is a three-digit whole number, $n$ must be divisible by 3 and be $\ge 100\cdot 3=300$ . If $3n$ is three digits, n must be $\le \frac{999}{3}=333$ So it must be divisible by three and between 300 and 333. There are $\boxed{\textbf{(A)}\ 12}$ such numbers, which you can find by direct counting.

Solution 2

Instead of finding n, we find $x=\frac{n}{3}$ . We want $x$ , $3x$ , and $9x$ to be three-digit integers. The smallest three-digit integer is $100$ , so that is our minimum value for $x$ , since if $x \in \mathbb{Z}$ , then $9x \in \mathbb{Z}$ . The largest three-digit integer divisible by $9$ is $999$ , so our maximum value is $\frac{999}{9}=111$ . There are $12$ numbers in the closed set $[\,100,111 ]\,$ , so the answer is $\boxed{\textbf{(A)}\ 12}$ .

- ColtsFan10

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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.

Page

Toolbox

Search

2008 AMC 8 Problems/Problem 22

Contents

Problem

Solution

Solution 2

See Also