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

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==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==
 
==Video Solution by OmegaLearn==

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

Video Solution 2

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


See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
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

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