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

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Solution 3

So we know the largest $3$ digit number is $999$ and the lowest is $100$ . This means $\dfrac{n}{3} \ge 100 \rightarrow n \ge 300$ but $3n \le 999 \rightarrow n \le 333$ . So we have the set ${300, 301, 302, \cdots, 330, 331, 332, 333}$ for $n$ . Now we have to find the multiples of $3$ suitable for $n$ , or else $\dfrac{n}{3}$ will be a decimal. Only numbers ${300, 303, \cdots, 333}$ are counted. We can divide by $3$ to make the difference $1$ again, getting ${100, 101 \cdots , 111}$ . Due to it being inclusive, we have $111-100+1 =\boxed{\textbf{(A) } 12}$

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=230

Video Solution 2

https://youtu.be/TFAp4X-OeE4 - Soo, DRMS, NM

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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