# 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 1

Instead of finding n, we find $x=\frac{n}{3}$. We want $x$ and $9x$ to be three-digit whole numbers. The smallest three-digit whole number 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 whole number divisible by $9$ is $999$, so our maximum value for $x$ is $\frac{999}{9}=111$. There are $12$ whole numbers in the closed set $\left[100,111\right]$ , so the answer is $\boxed{\textbf{(A)}\ 12}$.

- ColtsFan10

## Solution 2

We can set the following inequalities up to satisfy the conditions given by the question, $100 \leq \frac{n}{3} \leq 999$, and $100 \leq 3n \leq 999$. Once we simplify these and combine the restrictions, we get the inequality, $300 \leq n \leq 333$. Now we have to find all multiples of 3 in this range for $\frac{n}{3}$ to be an integer. We can compute this by setting $\frac{n} {3}=x$, where $x \in \mathbb{Z^+}$. Substituting $x$ for $n$ in the previous inequality, we get, $100 \leq x \leq 111$, and there are $111-100+1$ integers in this range giving us the answer, $\boxed{\textbf{(A)}\ 12}$.

- kn07

## Solution 3

We can create a list of the positive integers $n$ that fulfill the requirement of $\frac {n}{3}$ and $3n$ are three-digit whole numbers. The first number of this list must be $300$ since $\frac {300}{3} = 100$ is the smallest positive integer that satisfies this requirement. The last number of this list must be $333$ since $3 \cdot 333 = 999$ is the largest positive integer that satisfies this requirement. Since the problem requires $\frac {n}{3}$ and $3n$ must be whole numbers, the other numbers must be multiples of 3 (just like 300 and 333), so the list would look like this:

                                            $300, 303, 306, . . . , 333$


To put this list in to a countable form we must put it in a form similar to $1,2,3, . . ., n$. So, we manipulate it as follows:

                         $300-300,303-300,306-300, . . .,333-300 \Rightarrow 0,3,6, . . ., 33$

                         $\frac{0}{3}, \frac{3}{3}, \frac{6}{3}, . . ., \frac{33}{3} \Rightarrow 0,1,2, . . ., 11$

                              $0+1,1+1,1+2, . . ., 11+1 \Rightarrow 1,2,3, . . ., 12$


Now we can tell that there are 12 positive integers which satisfies the two requirements, so the answer is $\boxed{\textbf{(A)}\ 12}$.

~julia333