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

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==Solution 2==
 
==Solution 2==
Instead of finding n, we find <math>x=\frac{n}{3}</math>. We want <math>x</math>, <math>3x</math>, and <math>9x</math> to be three-digit integers. The smallest three-digit integer is <math>100</math>, so that is our minimum value for <math>x</math>, since if <math>x \in Z</math>, then <math>9x \in Z</math>. The largest three-digit integer divisible by <math>9</math> is <math>999</math>, so our maximum value is <math>\frac{999}{9}=111</math>. There are <math>12</math> numbers in the closed set <math>[\100,111]\$, so the answer is </math>\boxed{\textbf{(A)}\ 12}$.
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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>.
  
 
- ColtsFan10
 
- ColtsFan10

Revision as of 17:27, 28 August 2020

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

If $\frac{n}{3}$ is a three-digit whole number, $n$ must be divisible by 3 and be $\ge 100\cdot 3=300$. If $3n$ is three digits, n must be $\le \frac{999}{3}=333$ So it must be divisible by three and between 300 and 333. There are $\boxed{\textbf{(A)}\ 12}$ such numbers, which you can find by direct counting.

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

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
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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