Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 8 Problems/Problem 22"
(→‎Solution 4)
 
(19 intermediate revisions by 10 users not shown)
Line 8: Line 8:
 
\textbf{(E)}\ 34</math>
 
\textbf{(E)}\ 34</math>
  
−
==Video Solution==
+
==Solution 1==
−
https://youtu.be/rQUwNC0gqdg?t=230
 
−
 
 
−
==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
  
−
==Solution 3==
+
==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,
Line 24: Line 21:
 
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}
−
{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>.
−
-kn07
 
  
 +
- kn07
  
 +
==Solution 3==
  
 +
We can create a list of the positive integers <math>n</math> that fulfill the requirement of <math>\frac {n}{3}</math> and <math>3n</math> are three-digit whole numbers. The first number of this list must be <math>300</math> since <math>\frac {300}{3} = 100</math> is the smallest positive integer that satisfies this requirement. The last number of this list must be <math>333</math> since <math>3 \cdot 333 = 999</math> is the largest positive integer that satisfies this requirement. Since the problem requires <math>\frac {n}{3}</math> and <math>3n</math> must be whole numbers, the other numbers must be multiples of 3 (just like 300 and 333), so the list would look like this:
 +
                                            <math>300, 303, 306, . . . , 333</math>
 +
To put this list in to a countable form we must put it in a form similar to <math>1,2,3, . . ., n</math>. So, we manipulate it as follows:
 +
                          <math>300-300,303-300,306-300, . . .,333-300 \Rightarrow 0,3,6, . . ., 33</math>
  
 +
                          <math>\frac{0}{3}, \frac{3}{3}, \frac{6}{3}, . . ., \frac{33}{3} \Rightarrow 0,1,2, . . ., 11</math>
  
 +
                              <math>0+1,1+1,1+2, . . ., 11+1 \Rightarrow 1,2,3, . . ., 12</math>
 +
 +
Now we can tell that there are 12 positive integers which satisfies the two requirements, so the answer is <math>\boxed{\textbf{(A)}\ 12}</math>.
 +
 +
~julia333                 
 +
 +
==Solution 4==
 +
 +
We have to find the range for all numbers first. We can first find the smallest digit in which <math>\frac {n}{3}</math> is a 3 digit number that 3n is a 3 digit number which is <math>\frac {300}{3}</math> = 100, then we find the biggest number in which <math>\frac {n}{3}</math> is a 3 digit number and 3n is a 3 digit number which is <math>\frac {333}{3}</math> = 111. Now we know that all integers from 100 to 111 are possible values and the amount of integer values in that range is <math>\boxed{\textbf{(A)}\ 12}</math> (note I divided the numbers by 3 for convenience)
 +
 +
~twinotter
 +
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/rQUwNC0gqdg?t=230
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=21|num-a=23}}
 
{{AMC8 box|year=2008|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:34, 11 October 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

We can create a list of the positive integers $n$ that fulfill the requirement of $\frac {n}{3}$ and $3n$ are three-digit whole numbers. The first number of this list must be $300$ since $\frac {300}{3} = 100$ is the smallest positive integer that satisfies this requirement. The last number of this list must be $333$ since $3 \cdot 333 = 999$ is the largest positive integer that satisfies this requirement. Since the problem requires $\frac {n}{3}$ and $3n$ must be whole numbers, the other numbers must be multiples of 3 (just like 300 and 333), so the list would look like this: 

                                            $300, 303, 306, . . . , 333$

To put this list in to a countable form we must put it in a form similar to $1,2,3, . . ., n$. So, we manipulate it as follows: 

                         $300-300,303-300,306-300, . . .,333-300 \Rightarrow 0,3,6, . . ., 33$

                         $\frac{0}{3}, \frac{3}{3}, \frac{6}{3}, . . ., \frac{33}{3} \Rightarrow 0,1,2, . . ., 11$

                              $0+1,1+1,1+2, . . ., 11+1 \Rightarrow 1,2,3, . . ., 12$

Now we can tell that there are 12 positive integers which satisfies the two requirements, so the answer is $\boxed{\textbf{(A)}\ 12}$.

~julia333

Solution 4

We have to find the range for all numbers first. We can first find the smallest digit in which $\frac {n}{3}$ is a 3 digit number that 3n is a 3 digit number which is $\frac {300}{3}$ = 100, then we find the biggest number in which $\frac {n}{3}$ is a 3 digit number and 3n is a 3 digit number which is $\frac {333}{3}$ = 111. Now we know that all integers from 100 to 111 are possible values and the amount of integer values in that range is $\boxed{\textbf{(A)}\ 12}$ (note I divided the numbers by 3 for convenience)

~twinotter

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_22&oldid=229592"