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What is the smallest integer greater than 1 that leaves a remainder of 1 when divided by any of the integers 6, 8, and 10?
the answer is 121... how do you do this problem?
the answer is 121... how do you do this problem?
Since when did they teach this in middleschool? This seems soo hard, seems like Number theory or some crap...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.)
Mea Culpa.Originally posted by Hurkyl
Blarg, this is the second time (that I've noticed) that HallsofIvy has been accused of being me!