2008 AMC 8 Problems/Problem 22
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
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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 |
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