The discussion revolves around finding the smallest integer greater than 1 that leaves a remainder of 1 when divided by the integers 6, 8, and 10. Participants explore various methods and reasoning behind the problem, including algebraic approaches and the concept of least common multiples.
Participants express differing views on the complexity of the problem and its appropriateness for middle school education. While some agree on the methods to find the solution, there is no consensus on the classification of the problem or the level of difficulty it presents.
Some participants note that the problem requires whole number solutions, which complicates its classification within typical algebraic frameworks. There is also mention of the challenge posed by the problem, suggesting it may not be straightforward for all students.
Originally posted by HallsofIvy
Saying that the number, x, has remainder 1 when divided by 6, 8, and 10 means x= 6i+ 1, x= 8j+ 1, and x= 10k+ 1 for some i, j, k. Putting those together, 6i+ 1= 8j+ 1= 10k+ 1. Subtract 1 from each part: 6i= 8j= 10k. Divide each part by 2: 3i= 4j= 5k. Since 3, 4, and 5 are mutually prime, the smallest values that will work are i= 4*5, j= 3*5 and k= 3*4: in other words, 6i= 8j= 10k= 6*20= 4*15= 5*12= 120 and so x= 6i+ 1= 8j+ 1= 10k+1= 121.
Originally posted by NateTG
I know that lcm and gcf are covered in middle school, and that will get you the answer. (As I pointed out above.)
The horde of equations that Hurkyl used is more aptly described as basic algebra. (Although the exercise could certainly be described as a number theory problem.)
Originally posted by Hurkyl
Blarg, this is the second time (that I've noticed) that HallsofIvy has been accused of being me!