Difference between revisions of "2008 AMC 8 Problems/Problem 22"
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\textbf{(E)}\ 34</math> | \textbf{(E)}\ 34</math> | ||
| − | == | + | ==Solution 1== |
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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>. | 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 | ||
| − | ==Solution | + | ==Solution 2== |
We can set the following inequalities up to satisfy the conditions given by the question, | We can set the following inequalities up to satisfy the conditions given by the question, | ||
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- kn07 | - kn07 | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/rQUwNC0gqdg?t=230 | ||
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==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 21:14, 4 October 2022
Problem
For how many positive integer values of 
are both 
and 
three-digit whole numbers?
Solution 1
Instead of finding n, we find 
. We want 
and 
to be three-digit whole numbers. The smallest three-digit whole number is 
, so that is our minimum value for , since if 
, then 
. The largest three-digit whole number divisible by 
is 
, so our maximum value for is 
. There are 
whole numbers in the closed set ![$\left[100,111\right]$](http://latex.artofproblemsolving.com/a/8/d/a8da9aed381a9c6fbe90518505e54758d95ffe14.png)
, so the answer is 
.
- ColtsFan10
Solution 2
We can set the following inequalities up to satisfy the conditions given by the question,
,
and
.
Once we simplify these and combine the restrictions, we get the inequality, 
.
Now we have to find all multiples of 3 in this range for to be an integer. We can compute this by setting 
, where . Substituting for in the previous inequality, we get, 
, and there are 
integers in this range giving us the answer, .
- kn07
Video Solution
https://youtu.be/rQUwNC0gqdg?t=230
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. 
