2008 AMC 8 Problems/Problem 22

Revision as of 23:40, 29 August 2023 by Wiselion (talk | contribs) (Solution 3)

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 $111-100+1 =\boxed{(A) 12}$

Video Solution by OmegaLearn

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

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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