2015 AIME II Problems/Problem 1
Problem
Let be the least positive integer that is both percent less than one integer and percent greater than another integer. Find the remainder when is divided by .
Solution 1
If is percent less than one integer , then . In addition, is percent greater than another integer , so . Therefore, is divisible by 50 and is divisible by 25. Setting these two equal, we have . Multiplying by on both sides, we get .
The smallest integers and that satisfy this are and , so . The answer is .
Solution 2
Continuing from Solution 1, we have and . It follows that and . Both and have to be integers, so, in order for that to be true, has to cancel the denominators of both and . In other words, is a multiple of both and . That makes . The answer is .
Video Solution
https://www.youtube.com/watch?v=9re2qLzOKWk&t=7s
~MathProblemSolvingSkills.com
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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