2012 AMC 8 Problems/Problem 15
Contents
[hide]Problem
The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?
Video Solution
https://youtu.be/rQUwNC0gqdg?t=172
https://www.youtube.com/watch?v=Vfsb4nwvopU ~David
https://youtu.be/hOnw5UtBSqI ~savannahsolver
Solution
To find the answer to this problem, we need to find the least common multiple of , , , and add to the result. To calculate the least common multiple, we need to find the prime factorization for each of them. , , , and 2^{1}*{3^{1}}2^{2}*{3^{1}*{5^{1}}}, and by adding , we find that that such number is . Now we need to find the only given range that contains . The only such range is answer , and so our final answer is .
~NXC
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.