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Modulus and Congruency Problem

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data

    a) List all integers, A, that is in the range where A is greater than -51 and less than 51 such that it also satisfies: A is congruent to 7 (mod 17)

    b) has a set of representatives modulo 17, made up entirely of multiples of 3

    2. Relevant equations

    Only need to know what modulo is, which I believe is:
    when a is congruent to b (mod n)
    then (a-b) is a multiple integer of n

    also, mod is for finding the remainder of a division

    3. The attempt at a solution
    (x-7)=17n ---> x = 17n + 7
    a) going by what I think is true (above) then I think possible answers for A are:
    -44,-27,-10,7,24,41

    < never done congruencies before >
    Is that right?

    b) This is where I got really stumped because it seemed too easy:
    0,3,6,9,12,15,18,21,24,27,30, 33,36,39,42,45,48 (similarly for the negatives) ???
     
  2. jcsd
  3. Jan 17, 2010 #2

    Mark44

    Staff: Mentor

    For a) you're not far off. The set is {-41, -24, -7, 10, 27, 44}. Notice that this set is {-34 - 7, -17 - 7, 0 - 7, 17 - 7, 34 - 7, 51 - 7}.

    For b) I'm not totally clear what the problem is asking for. Is it the numbers in the first set that are multiples of 3? If so, only -24 and 27 would be included.
     
  4. Jan 17, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Good. Notice that each of those is the previous number plus 17.

    No, it is the remainders, after division by 17 that must be multiples of 3:
    3, 20, 37: 3= 0(17)+ 3, 20= 1(17)+ 3, 37= 2(17)+ 3. Again, those differ by 17. I started with 3 and added 17. To get the negatives, subtract 17 rather than add: 3- 13= -14, -31, -48: -14= -1(17)+ 3, -31= -2(17)+ 3, -48= -3(17)+ 3
     
  5. Jan 19, 2010 #4
    Why is it the remainders after division by 17, and not before, that has to be multiples of 3?
     
    Last edited: Jan 19, 2010
  6. Jan 19, 2010 #5

    Mark44

    Staff: Mentor

    Because of the way the problem is stated.
    You want the numbers in this set -- {-41, -24, -7, 10, 27, 44} -- that are divisible by 3. At least that was my interpretation of the problem. I asked for clarification in post #2, but you didn't reply to my question.
     
  7. Jan 19, 2010 #6
    whoups! Part b does not refer to the previous set in part a. Part b is asking for a completely new set.
     
  8. Jan 20, 2010 #7

    Mark44

    Staff: Mentor

    The representatives mod 17 are the numbers in the set {0, 1, 2, 3, ..., 15, 16}. Which of them are multiples of 3?
     
  9. Jan 20, 2010 #8

    HallsofIvy

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    What could you possibly mean by remainder before dividing? There are no remainders until you have divided!
     
  10. Jan 23, 2010 #9
    obviously 3, 6, 9, etc... as I originally thought, but that would be almost too easy. Would this set "skip" over number 51 (which is 17x3) and 102 (17x6), ... etc? Should this set just be listed from 3 to 48? or can I just list 3, 6, 9 and then put a "..." and assume others will know what I meant? The question never asked for it to be a finite set.

    I asked that because of the way the question was worded, and I realized I worded my question very poorly (sorry); I wasn't sure if it was asking for a number mod 17 that was divisible by 3 or a number divisible by 3 mod 17.
     
  11. Jan 23, 2010 #10

    Mark44

    Staff: Mentor

    If I understand the second problem (I'm not sure that I do), what you want are the representatives modulo 17 that are multiples of 3. The representatives mod 17 are 0, 1, 2, 3, 4, ..., 15, and 16. The answer to the question as I understand it are 3, 6, 9, 12, and 15.
     
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