2017 AMC 8 Problems/Problem 12
Contents
[hide]Problem
The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?
Solution 1
Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. The is . Since , that is in the range of
Solution 2
Call the number we want to find . We can say that We can also say that is divisible by , and Therefore, , so which is in the range of
~PEKKA
Video Solution (CREATIVE THINKING!!!)
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Video Solution
~savannahsolver
https://www.youtube.com/watch?v=nOSqQjEv0U0 ~David
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AJHSME/AMC 8 Problems and Solutions |
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