2008 AMC 8 Problems/Problem 22

Revision as of 17:27, 28 August 2020 by Coltsfan10 (talk | contribs) (Solution 2)

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 $x$ 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

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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