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MCIII NT Chap1 HW help
Hi! I've beeen working through the practice problems of MCIII NT Chapter 1, and I've encountered a few problems I'm stuck on. On page 29-30, 1.2 Practice Questions, I have trouble solving Problem 1.14-1.20. I would really appreciate any hints or solutions.
For 1.14 I'm trying to prove by contradiction but I don't know where to start
For 1.15 I've used n to express an equation but am stuck on that
For 1.16 I think I've solved most of the problem but I'm not sure about the part where m isn't prime, for this problem I would truly appreciate a solution.
For 1.17-1.20 I'm a little dumbfounded and don't know where to start.
I have looked through the answer key but most were omitted. Any hints, solutions and help would be greatly appreciated. Thank you!
P.S. If I submitted my writing problems after the HW is due, will a teacher still grade my written out solution?
For 1.14, try to use Bezout's Identity.
For 1.15, if you have an equation, that's the way to go. You can show your equation here and ask further. On the other hand, if you just try to calculate the digits one at a time, starting from the units digit, you can get it too, without using any equation.
For 1.16, try a few examples of $m$. If you can solve the case where $m$ is a prime, then the non-prime case is not too different. Show how far you have got here, and you will get further hints.
For 1.17, try to add all values in a complete residue system. Again, an example can help you.
For 1.18: this is easier. No hint is needed. Think a little more.
1.19 and 1.20 require some in-depth analysis. For 1.19 you can try to consider cases where $n$ is even and $n$ is odd.
For 2.20, start with the first digit. What is $a$? Then think about the following questions: Can the digits repeat? What can be the last digit? Narrow down the choices.
For 1.15, my equation was: 2x10^a+N=20N+4, a being the number of digits of N. I got stuck there unfortunately.
For 1.16, I kind of have a solution but I'm not sure if it's correct.
if m is prime, then k and m-k are add to 0 modulo m, and there can't be one in the middle because that way 2 would divide m.
if m isn't, if a is part of its reduced residue system, then m-a must be as well. and we can pair them up.
1.15 just need a little more thought. You can transform the equation to $2\times 10^a = 19N + 4$. Think more carefully what this equation means. Hint: what do you do if you divide 200 by 19? 2000 by 19? 20000 by 19?
1.16 is correct. Pairing is a good method that can be applied to many problems. Good job!
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