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 and is divisible by . 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 .
Solution 3
We can express as and , where and are some positive integers. Let us try to find the smallest possible value of , first ignoring the integral constraint.
Obviously, we are just trying to find the LCM of and They have no common factors but , so we multiply and and divide by to get This is obviously not divisible by , so this is not possible, as it would imply that which is not an integer. This can be simplified to .
We know that any possible value of will be an integral multiple of this value; the smallest such is achieved by multiplying this value by We arrive at , which is
~Technodoggo
Video Solution
https://www.youtube.com/watch?v=9re2qLzOKWk&t=7s
~MathProblemSolvingSkills.com
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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