2008 AMC 8 Problems/Problem 22
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
We can create a list of the positive integers that fulfill the requirement of and are three-digit whole numbers. The first number of this list must be since is the smallest positive integer that satisfies this requirement. The last number of this list must be since is the largest positive integer that satisfies this requirement. Since the problem requires and must be whole numbers, the other numbers must be multiples of 3 (just like 300 and 333), so the list would look like this:
To put this list in to a countable form we must put it in a form similar to . So, we manipulate it as follows:
Now we can tell that there are 12 positive integers which satisfies the two requirements, so the answer is .
~julia333
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 | |
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