Difference between revisions of "2008 AMC 8 Problems/Problem 22"
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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>. | ||
Revision as of 16:04, 23 March 2023
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
, 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 by OmegaLearn
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.