Solve the congruence 24x �identical to 42 (mod 30) for x. List all the solutions (if there are any).
24x = 42 mod 30
24x = 12 mod 30
(24-30)x = 12 mod 30
-6x = 12 mod 30
x = -2
so one possible solution is x = -2(24) = -48
adding/subtracting multiples of 30 , our mod, will yield other solutions
making -2(24) + 60 , or 12 , our smallest positive answer
A general solution would be 12 + 30k, where k is an integer
check:
look at the sequence of numbers
... -108 - 78 - 48 -18 12 42 72 102 ...
IF we take 102 ÷ 30 we get
3 + 12/30
or
2 + 42/30
so clearly we can see that 102 = 12 mod 30 = 42 mod 30
To solve the congruence 24x ≡ 42 (mod 30) for x, we need to find the values of x that satisfy the equation.
Step 1: Determine the greatest common divisor (GCD) of 24 and 30.
GCD(24, 30) = 6
Step 2: Check if 42 is divisible by the GCD obtained in Step 1.
42 is divisible by 6, so the congruence is solvable.
Step 3: Divide both sides of the congruence by the GCD.
Dividing both sides by 6, we have:
4x ≡ 7 (mod 5)
Step 4: Find the modular inverse of 4 (mod 5). The modular inverse of a number a (mod n) is the number b such that (a * b) ≡ 1 (mod n).
To find the modular inverse of 4 (mod 5), we can use the extended Euclidean algorithm:
5 = 4 * 1 + 1
1 = 5 - 4 * 1
Therefore, the modular inverse of 4 (mod 5) is 1.
Step 5: Multiply both sides of the congruence by the modular inverse.
Multiplying both sides by 1 (the modular inverse of 4), we have:
1 * 4x ≡ 1 * 7 (mod 5)
4x ≡ 7 (mod 5)
Step 6: Simplify the congruence.
Since 4 * 4 ≡ 1 (mod 5), we can multiply both sides of the congruence by 4:
4 * 4x ≡ 4 * 7 (mod 5)
16x ≡ 28 (mod 5)
Step 7: Reduce the coefficients and the constant in the congruence.
16x ≡ 3 (mod 5)
x ≡ 3 (mod 5)
Step 8: Find the solution for x.
The solutions for x can be found by adding any multiple of 5 to the congruence:
x = 3 + 5k
where k is an integer.
Therefore, the solutions for x are: 3, 8, 13, 18, 23, 28, ...
So, x can take on any value from this infinite sequence.