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Number Theory: Congruences

  1. Sep 21, 2010 #1
    I am on the http://cow.temple.edu/~cow/cgi-bin/manager website working some congruence problems, here you can plug in answers over and over until you get them right.

    Three problems still baffle me:

    1) With Mod24, find the solution of 3-15-21=. Here I just pretended that none of the numbers were negative (which meant they totaled 39) and arrived at an answer of 15, which was correct.

    2) With Mod25, find the solution of 3-5+13-24=. I tried the same approach of pretending that none of the numbers were negative, and got the wrong answer. I then just added with each sign as it was (to get a total of -13) and arrived at a correct solution of 12.

    What is the difference between these two problems that changes the methods of solving them, or am I going about it totally wrong? The earlier addition/subtraction problems seemed straightforward.

    3) With Mod 24^x, find the solution of the inverse of 17. I set the problem up as 17x=1Mod24.
    As hard as I tried (I used every number from 1 to 23), I could not get an answer to work out. I just randomly plugged 9 in as the solution and it was the answer. When I went back and plugged 9 into my equation it did not match up at all. I have been successful at doing many of these inverse type problems, but I just do not understand this one. The only thing I could get to equal 9 was the number of prime numbers in 24. What am I misunderstanding?

    Thanks for the help
     
  2. jcsd
  3. Sep 23, 2010 #2
    In your first question, the second way you did it with mod 25 is correct. When you got right the first one by switching the signs, it was a coincidence. Note that 3-15-21 = 3-(15+21) = 3-12 (mod 24). However, add 24, and you find that 3-12 = 3+12 = 3+(15+21) (mod 24). This only worked because 15+26 = 12 (mod 24). Of course, 25 is odd, so the same coincidence couldn't happen.

    For your other question, I really don't know. When you say mod 24^x, are you supposed to solve for x? And in fact 17 does have an inverse mod 24, namely itself.
     
  4. Sep 23, 2010 #3
    On problem 1 (Mod 24 3-15-21=x) ,why did you reverse the sign on the 21 (3-(15+21)) to find your solution of 3-12Mod24?

    Thanks for your help, math has always been a trouble spot for me, but I feel it is the purest form of logic, so I want to get good at it.
     
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