# 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

- Wayne DeguManLv 72 months agoFavorite 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 72 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 HLv 72 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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- PopeLv 72 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