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

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:

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:

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

@@ Line 8: / Line 8: @@
 \textbf{(E)}\ 34</math>
-==Video Solution==
+==Solution 1==
-https://youtu.be/rQUwNC0gqdg?t=230
+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
 ==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>.
+We can set the following inequalities up to satisfy the conditions given by the question,
+<math>100 \leq \frac{n}{3} \leq 999</math>,
+and
+<math>100 \leq 3n \leq 999</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}
+{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
+==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
-- ColtsFan10
+==Video Solution by OmegaLearn==
+https://youtu.be/rQUwNC0gqdg?t=230
 ==See Also==
 {{AMC8 box|year=2008|num-b=21|num-a=23}}
 {{MAA Notice}}

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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