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Math Help - Modulo problem

  1. #1
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    Post Modulo problem

    Hello guys new here,

    i really cant understand this topic which my professor taught us. It was about modulus. At first i can understand what hes been teaching however i cant get this part about modulus. I learned that modulus is simply the remainder. I also learned how to construct tables from a particular modulo. For example:,

    Constructing table of Z(3)

    Z(3) = {0,1,2}



    however what i cant understand is this, he said that in Z(3):

    -1=additive inverse of 1=2
    -2=additive inverse of 2=1

    1/2=multiplicative inverse of 2=2


    According to my research in the internet the additive inverse of a number is a number which when you add to that number will be equal to 0. So how come that in modulo of 3, the additive inverse of 1=2?

    For example my professor had given us this problem:

    In Z(3) find the ff:

    a)-2-1

    Solution: -2-1=1+2=3=0

    I will appreciate it greatly if someone can help me thanks in advance
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  2. #2
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    In Z(3), the only acceptable numbers are {0, 1, 2}.

    You want to know what number adds with 1 to get 0. This gives you the additive inverse.

    Look on your additive table, what adds with 1 to get 0?
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  3. #3
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    wow great thanks !
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  4. #4
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    Another problem of mine - How to compute for residue and unit digit

    Sorry to bother you again guys but time is running out for me and surely if i take my time for understanding this i will be not able to review for other subjects so please guys really need your help.

    My last problem is how to find for the least residue and the unit digit? If youre not busy please add some brief explanations on how to do the process. Thanks thanks hope to help you guys in the future too =D


    How to find the least residue of 37^13(mod17)?

    How to find the unit digit of 3^103?
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  5. #5
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    How to find the least residue of 37^13(mod17)?
    Use a big number calculator, like this Java applet .

    The fact is, computing the remainder is interchangeable with computing arithmetic operations. I.e., if you want to find the remainder of x+y over z, you can first add x and y and then find the remainder, or you can find the remainder of x, the remainder of y, add them, and find the remainder again of the result is bigger than z. A similar thing goes for multiplication.

    So, since 37 = 3 (mod 17), instead of computing 37^13 mod 17, you can find 3^13 mod 17. Next, 3^3 = 27 = 10 (mod 17), so 3^6 = (3^3)^2 = 10 ^ 2 = 100 = 15 = -2 (mod 17). Therefore, 3^12 = (3^6)^2 = (-2)^2 = 4 (mod 17), and 3^13 = 4*3 (mod 17).

    How to find the unit digit of 3^103?
    Note that as you compute the successive powers of 3, the last digit changes as follows: 1 (for 3^0) -> 3 -> 9 -> 7 -> 1. So you only need to figure out which of those 4 digits will stand in the 103th place.
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  6. #6
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    guys need some help in my subject, im really confused bout this part coz im absent when my professor taught this to my section,

    ....according to Modulo Congruence it is correct to say that 38 ≡ 14 (mod 12) and also 2 ≡ 14 (mod 12)

    hmm it is clear to me that when you divide 38 and 14 by 12 the remainder is 2 so the congruence for that is true. However the remainder of 14/12 is 2 while 2/12 is a fraction so how come that 2 is congruent to 14 modulo 12.

    Thanks guys really nid your help
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  7. #7
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    You don't deal with fractions in this topic. 2 = 0 * 12 + 2, so the ratio is 0 and the remainder is 2.
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  8. #8
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    Quote Originally Posted by Neverwinter View Post
    guys need some help in my subject, im really confused bout this part coz im absent when my professor taught this to my section,

    ....according to Modulo Congruence it is correct to say that 38 ≡ 14 (mod 12) and also 2 ≡ 14 (mod 12)

    hmm it is clear to me that when you divide 38 and 14 by 12 the remainder is 2 so the congruence for that is true. However the remainder of 14/12 is 2 while 2/12 is a fraction so how come that 2 is congruent to 14 modulo 12.

    Thanks guys really nid your help
    Don't know if this will help you but

    a-b)" alt=" a\equiv b\pmod{m} \iff m\:|\a-b)" />

    I don't suppose you've learned what an equivalence class is?

    Also I learned that the proper term is common residue, not sure how standard least residue is.
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  9. #9
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    However the remainder of 14/12 is 2 while 2/12 is a fraction so how come that 2 is congruent to 14 modulo 12.


    hmm this is what mostly bothers me
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  10. #10
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    Quote Originally Posted by Neverwinter View Post

    hmm this is what mostly bothers me
    The multiples of 12 are {..., -24, -12, 0, 12, 24, ...}

    2\equiv14\pmod{12} means that 2-14=-12 is divisible by 12, ie, is a multiple of 12.

    We're not really dividing and getting a remainder, that's a simplified explanation that applies under certain conditions.

    For example it's also true that 14\equiv26\pmod{12}.
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