Difference between revisions of "2008 AMC 8 Problems/Problem 22"

Revision as of 08:37, 20 June 2024

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

Video Solution by OmegaLearn

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

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.

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 - kn07
-==
-==Solution 3==
-So we know the largest <math>3</math> digit number is <math>999</math> and the lowest is <math>100</math>. This means <math>\dfrac{n}{3} \ge 100 \rightarrow n \ge 300</math> but <math>3n \le 999 \rightarrow n \le 333</math>. So we have the set <math>{300, 301, 302, \cdots, 330, 331, 332, 333}</math> for <math>n</math>. Now we have to find the multiples of <math>3</math> suitable for <math>n</math>, or else <math>\dfrac{n}{3}</math> will be a decimal. Only numbers <math>{300, 303, \cdots, 333}</math> are counted. We can  divide by <math>3</math> to make the difference <math>1</math> again, getting <math>{100, 101 \cdots , 111}</math>. Due to it being inclusive, we have <math>111-100+1 =\boxed{\textbf{(A)} 12}</math>
 ==Video Solution by OmegaLearn==

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