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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]\,</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>, <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]\,</math>, so the answer is <math>\boxed{\textbf{(A)}\ 12}</math>.
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- ColtsFan10
  
 
==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}}

Revision as of 19:23, 21 March 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 $x$, $3x$, and $9x$ to be three-digit integers. The smallest three-digit integer is $100$, so that is our minimum value for $x$, since if $x \in Z$, then $9x \in Z$. The largest three-digit integer divisible by $9$ is $999$, so our maximum value is $\frac{999}{9}=111$. There are $12$ numbers in the closed set $[\,100,111]\,$, so the answer is $\boxed{\textbf{(A)}\ 12}$.

- ColtsFan10

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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