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Difference between revisions of "2008 AMC 8 Problems/Problem 22"
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\textbf{(E)}\ 34</math>
 
\textbf{(E)}\ 34</math>
  
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==Video Solution==
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==Solution 1==
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https://youtu.be/rQUwNC0gqdg?t=230
 
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==Solution 2==
 
 
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>.
  
 
- ColtsFan10
 
- ColtsFan10
  
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==Solution 3==
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==Solution 2==
 
   
 
   
 
We can set the following inequalities up to satisfy the conditions given by the question,
 
We can set the following inequalities up to satisfy the conditions given by the question,
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Once we simplify these and combine the restrictions, we get the inequality, <math>300 \leq n \leq 333</math>.
 
Once we simplify these and combine the restrictions, we get the inequality, <math>300 \leq n \leq 333</math>.
 
Now we have to find all multiples of 3 in this range for <math>\frac{n}{3}</math> to be an integer. We can compute this by setting <math>\frac{n}
 
Now we have to find all multiples of 3 in this range for <math>\frac{n}{3}</math> to be an integer. We can compute this by setting <math>\frac{n}
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{3}=x</math>, where <math>x \in \mathbb{Z^+}</math>. Substituting <math>x</math> for <math>n</math> in this inequality, we get, <math>100 \leq x \leq 111</math>, and there are <math>111-100+1</math> integers in this range giving us the answer, <math>\boxed{\textbf{(A)}\ 12}</math>.
+
{3}=x</math>, where <math>x \in \mathbb{Z^+}</math>. Substituting <math>x</math> for <math>n</math> in the previous inequality, we get, <math>100 \leq x \leq 111</math>, and there are <math>111-100+1</math> integers in this range giving us the answer, <math>\boxed{\textbf{(A)}\ 12}</math>.
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-kn07
 
  
 +
- kn07
  
  
  
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 +
==Solution 3==
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 +
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==
 +
https://youtu.be/rQUwNC0gqdg?t=230
 +
 +
==Video Solution 2==
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https://youtu.be/TFAp4X-OeE4 - Soo, DRMS, NM
  
  

Latest revision as of 20:57, 26 January 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



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

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 (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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