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 agoFavourite 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