2008 AMC 8 Problems/Problem 22: Difference between revisions

Revision as of 18:42, 4 October 2022

Problem

For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$

Video Solution

Solution 2

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 3

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 this 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 15: / Line 15: @@
 - ColtsFan10
+==Solution 3==
+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 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>.
+-kn07
 ==See Also==
 {{AMC8 box|year=2008|num-b=21|num-a=23}}
 {{MAA Notice}}

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