Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Find all the whole number solutions of congruence equation. Pls answer any of the three pls?

1.) 3x + 12 ≡ 7 mod 10

2.) 4x + 6 ≡ 5 mod 8

3.) 5x + 3 ≡ 8 mod 12

6 Answers

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  • ?
    Lv 7
    2 months ago
    Favourite answer

    1) 3x ≡ -5 mod10

    i.e. 3x ≡ 5 mod10

    Multiplying by 7 we get:

    21x ≡ 35 mod10

    i.e. x ≡ 5 mod10 

    Hence, x = 10k + 5

    2) 4x ≡ -1 mod8

    i.e. 4x ≡ 7 mod8

    No solutions exist as the g.c.d. of 4 and 8,...i.e. 4, is not a divisor of 7

    3) 5x ≡ 5 mod12

    Multiplying by 5 we get:

    25x ≡ 25 mod12

    i.e. x = 1 mod12

    Hence, x = 12k + 1

    :)>

  • ?
    Lv 7
    2 months ago

     3x + 12 ≡ 7 mod 10

     x congruent 5 (mod 10)

  • 2 months ago

    3x+12=7(mod 10)

    =>

    3x=-5(mod 10)

    =>

    3x=5(mod 10)

    =>

    x=5(mod 10)

    =>

    x=5+10k, where k=an integer.

    is the G.S.

    4x+6=5(mod 10)

    =>

    4x+1=0(mod 10)

    (4x+1)/10=/= an integer

    since 4x+1 is always an odd number.

    =>

    there is no solution of x.

    5x+3=8(mod 12)

    =>

    5x=5(mod 12)

    =>

    x=1(mod 12)

    =>

    x=1+12k, where k= an integer

    is the G.S.

  • Ian H
    Lv 7
    2 months ago

    5x + 3 ≡ 8 mod 12

    5x = 5 mod 12

    x = 12 + 1 is one solution, because

    5(12 + 1) = 0 mod 12 + 5 mod 12

    x = 24 + 1 is another solution

    x = 12n + 1 is a general solution

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  • 2 months ago

    The answer is 42.  

  • Pope
    Lv 7
    2 months ago

    3x + 12 ≡ 7 (mod 10)

    Begin with 12 and add multiples of 3 until you get a number with 7 as the unit digit. The first one arises with 5.

    3(5) + 12 = 27

    3(5) + 12 ≡ 7 (mod 10)

    At this point adding 3 times any positive multiple of 10 also will result in an integer having 7 as the unit digit.

    x = 5 + 10k, for any non-negative integer k

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