Thread: Division algorithm/mod

Find the smallest positive integer that has remainder 2 when divided by 5, remainder 3 when divided by 7, remainder 4 when divided by 9, and remainder 5 when divided by 11. Hint: Solve 2+5k=3(mod 7) to start, and continue.

Solving 2+5k=3(mod 7) I think gives that 7 divides 5k-1, and so, 7 divides 5k+6, but I don't know how that helps or where to go from here.

I got the answer is 1732 by the Chinese Remainder Theorem, but we can't use that since we haven't shown it in class, and so, I'm still confused why we would solve 2+5k=3(mod 7), and what equations we solve next using the method suggested.

Using the Chinese Remainder Theorem seems to be the best approach to solving this problem. I'm confused as to why you are being asked a question that begs to be solved using a theorem you can't use.

I don't know. It just says we should solve 2+5k=3(mod 7), and to continue in this fashion to get the result. We haven't covered the Chinese Remainder Theorem in this course, and I know he is the kind of prof to deduct points for not doing things his way. (it is actually for abstract alg, and so, we do not cover it in this course).