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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Once we simplify these and combine the restrictions, we get the inequality, <math>300 \leq n \leq 333</math>. | Once we simplify these and combine the restrictions, we get the inequality, <math>300 \leq n \leq 333</math>. | ||
Now we have to find all multiples of 3 in this range for <math>\frac{n}{3}</math> to be an integer. We can compute this by setting <math>\frac{n} | Now we have to find all multiples of 3 in this range for <math>\frac{n}{3}</math> to be an integer. We can compute this by setting <math>\frac{n} | ||
− | {3}=x</math>, where <math>x \in \mathbb{Z^+}</math>. Substituting <math>x</math> for <math>n</math> in | + | {3}=x</math>, where <math>x \in \mathbb{Z^+}</math>. Substituting <math>x</math> for <math>n</math> in the previous inequality, we get, <math>100 \leq x \leq 111</math>, and there are <math>111-100+1</math> integers in this range giving us the answer, <math>\boxed{\textbf{(A)}\ 12}</math>. |
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+ | - kn07 | ||
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+ | ==Solution 3== | ||
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+ | So we know the largest <math>3</math> digit number is <math>999</math> and the lowest is <math>100</math>. This means <math>\dfrac{n}{3} \ge 100 \rightarrow n \ge 300</math> but <math>3n \le 999 \rightarrow n \le 333</math>. So we have the set <math>{300, 301, 302, \cdots, 330, 331, 332, 333}</math> for <math>n</math>. Now we have to find the multiples of <math>3</math> suitable for <math>n</math>, or else <math>\dfrac{n}{3}</math> will be a decimal. Only numbers <math>{300, 303, \cdots, 333}</math> are counted. We can divide by <math>3</math> to make the difference <math>1</math> again, getting <math>{100, 101 \cdots , 111}</math>. Due to it being inclusive, we have <math>111-100+1 =\boxed{\textbf{(A) } 12}</math> | ||
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+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/rQUwNC0gqdg?t=230 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/TFAp4X-OeE4 - Soo, DRMS, NM | ||
Revision as of 20:57, 26 January 2024
Contents
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
Solution 3
So we know the largest digit number is and the lowest is . This means but . So we have the set for . Now we have to find the multiples of suitable for , or else will be a decimal. Only numbers are counted. We can divide by to make the difference again, getting . Due to it being inclusive, we have
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=230
Video Solution 2
https://youtu.be/TFAp4X-OeE4 - Soo, DRMS, NM
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.