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>[\ | + | 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}$. |
- ColtsFan10 | - ColtsFan10 | ||
Revision as of 19:23, 21 March 2020
Contents
Problem
For how many positive integer values of 
are both 
and 
three-digit whole numbers?
Solution
If is a three-digit whole number, must be divisible by 3 and be 
. If is three digits, n must be 
So it must be divisible by three and between 300 and 333. There are 
such numbers, which you can find by direct counting.
Solution 2
Instead of finding n, we find 
. We want 
, 
, and 
to be three-digit integers. The smallest three-digit integer is 
, so that is our minimum value for , since if 
, then 
. The largest three-digit integer divisible by 
is 
, so our maximum value is 
. There are 
numbers in the closed set $[\100,111]$, so the answer is$ (Error compiling LaTeX. Unknown error_msg)\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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
