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

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Instead of finding n, we find <math>x=\frac{n}{3}</math>. We want <math>x</math> and <math>9x</math> to be three-digit whole numbers. The smallest three-digit whole number is <math>100</math>, so that is our minimum value for <math>x</math>, since if <math>x \in \mathbb{Z^+}</math>, then <math>9x \in \mathbb{Z^+}</math>. The largest three-digit whole number divisible by <math>9</math> is <math>999</math>, so our maximum value for <math>x</math> is <math>\frac{999}{9}=111</math>. There are <math>12</math> whole numbers in the closed set <math>\left[100,111\right]</math> , so the answer is <math>\boxed{\textbf{(A)}\ 12}</math>.
 
Instead of finding n, we find <math>x=\frac{n}{3}</math>. We want <math>x</math> and <math>9x</math> to be three-digit whole numbers. The smallest three-digit whole number is <math>100</math>, so that is our minimum value for <math>x</math>, since if <math>x \in \mathbb{Z^+}</math>, then <math>9x \in \mathbb{Z^+}</math>. The largest three-digit whole number divisible by <math>9</math> is <math>999</math>, so our maximum value for <math>x</math> is <math>\frac{999}{9}=111</math>. There are <math>12</math> whole numbers in the closed set <math>\left[100,111\right]</math> , so the answer is <math>\boxed{\textbf{(A)}\ 12}</math>.
  

Revision as of 15:04, 23 March 2023

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

See Also

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

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