Question #254916

Solve the congruence 5x≡1(mod12)

Hint: 0≤x≤11


1
Expert's answer
2021-12-21T03:14:01-0500

axb(mod m)ax ≡ b(mod \space m)

Step 1

GCD of a and m

GCD of 5 & 12 

since 5 is a prime , so the GCD is 1

step 2

1/1=1

So, solutions exists

step 3

5x1+12n(mod 12)5x ≡1+12n(mod \space 12)

where n is a natural number

finding n such that (1+12n) is divided by 5

if n= 1

5x1+12(mod 12)5x13(mod 12)5x ≡1+12(mod \space 12)\\ 5x ≡13(mod \space 12)

for n=2

5x1+122(mod 12)5x25(mod 12)x5(mod 12)5x ≡1+12*2(mod \space 12)\\ 5x ≡25(mod \space 12)\\ x ≡5(mod \space 12)

So, the solution of n between 0 and 11 will be 5. In general form it will be x = 5+12m


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