Number Theory - congruence problem

[Jamers328]

36x = 8 (mod 102).


I just need to find x = ____ (mod 102).

I know I have to manipulate 36 and 8 by adding multiples of 102, or multiplying by the same number, but I'm not getting anywhere.

Thanks.

 

[arthur ohlsten]

36x=8 mod 120
I do not believe this has an answer.

36x=8+ 120n
x=8/36 + 120n/36
x=2/9 + [120/36]n
x=2/9 + [3+12/36]n
x=2/9 + [3+3/9]n
x=3n + [3n+2]/9

is there an integer such that 3n+2 is a multiple of 9? no
n....... remainder
1.............5
2.............8
3.............2
4.............5
5.............8

the remainder is always 2,5, or 8

Arthur

 

[Jamers328]

I agree. Thank you!!!!

 

[Deleted member 4993]

36x=8 mod 120
I do not believe this has an answer.



is there an integer such that 3n+2 is a multiple of 9?

No - because (3n+2) is not divisible by 3 [neither is (3n+1)]

Arthur

 

[daon]

The modulus was 102, not 120. Either way the answer is the same.

 

[arthur ohlsten]

Daon-
Sorry for the error
You are right it still has no answe.
By the same method the remainders are
n.....remainder
1..........1
2..........8
3..........13
4..........10
5..........7
6..........4

36x=8 mod 102
x= 8/36 +102 n /36
x=2/9 +[2+30/36]n
x= 2n+2/9 + 5n/6
x=2n + [4+15n]18

Arthur