Consider the following equation in
𝑍
27
:

3
𝑥
=
7

How many solutions exist for
𝑥
?

Question

Consider the following equation in 
𝑍
27
:

3
𝑥
=
7

How many solutions exist for 
𝑥
?

BlackTom AI · Accepted Answer

We are given the congruence 3x ≡ 7 (mod 27) and asked how many solutions x in Z27 exist.
First, recall a standard result for linear congruences: the equation a x ≡ b (mod m) has solutions iff gcd(a, m) divides b.
Here, a = 3 and m = 27, so gcd(3, 27) = 3.
We check whether 3 divides b = 7. Since 7 modulo 3 is 1, 7 is not divisible by 3.
Because gcd(3, 27) = 3 does not divide 7, the congruence 3x ≡ 7 (mod 27) has no solutions in integers modulo 27.
Intuitively, multiplying any residue class by 3 mod 27 can only yield multiples of 3 mod 27, but 7 is not a multiple of 3, so no residue x can produce 7 when multiplied by 3 under modulo 27.
If one wanted to verify directly, you could attempt to solve the congruence by reducing both sides modulo gcd(3,27); since 7 is not a multiple of 3, there is a contradiction, confirming the lack of solutions.
Thus, in the ring Z27, this equation has zero solutions, and there are no x in Z27 satisfying 3x ≡ 7 (mod 27).

MATH3066 (ND) 2025 Mid-Semester Test Practice Set C

Consider the following equation in 𝑍 27 : 3 𝑥 = 7 How many solutions exist for 𝑥 ?

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