Difference between revisions of "2008 AMC 8 Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | If <math>\frac{n}{3}</math> is a three-digit whole number, <math>n</math> must be divisible by 3 and be <math>\ge 100\cdot 3=300</math>. If <math>3n</math> is three digits, n must be <math>\le \frac{999}{3}=333</math> So it must be divisible by three and between 300 and 333. There are <math>\boxed{\textbf{(A)}\ 12}</math> such numbers, which you can find by direct counting. | + | If <math>\frac{n}{3}</math> is a three-digit whole number, <math>n</math> must be divisible by 3 and be <math>\ge 100\cdot 3=300</math>. If <math>3n</math> is three digits, n must be <math>\le \frac{999}{3}=333</math>. So it must be divisible by three and between 300 and 333. There are <math>\boxed{\textbf{(A)}\ 12}</math> such numbers, which you can find by direct counting. |
==Solution 2== | ==Solution 2== |
Revision as of 19:02, 14 August 2022
Contents
[hide]Problem
For how many positive integer values of are both and three-digit whole numbers?
Video Solution
https://youtu.be/rQUwNC0gqdg?t=230
Solution
If is a three-digit whole number, must be divisible by 3 and be . If is three digits, n must be . So it must be divisible by three and between 300 and 333. There are such numbers, which you can find by direct counting.
Solution 2
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
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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