Let Σ={1,2,3,4} For x ∈ Σ*, let prod(x) be the product of symbols in x modulo 7. We take prod(ϵ)=1, where ϵ is the null string. For example, prod(124)=(1×2×4) mod  7=1 prod(124)=(1×2×4)mod7=1. Define L={x∈Σ∗ ∣ prod(x)=2}. The number of states in a minimum state DFA for L is ______

The correct answer is 6.

We are given the alphabet $\Sigma =\{1,2,3,4\}$.
For any string $x\in {\Sigma }^{*}$, $\mathrm{prod}\left(x\right)$ is defined as the product of the symbols in $x$ modulo 7, with $\mathrm{prod}\left(ϵ\right)=1$.
We want to find the number of states in the minimum state Deterministic Finite Automaton (DFA) that accepts the language:
$L=\{x\in {\Sigma }^{*}\mid \mathrm{prod}(x)=2\}$.

Let's analyze the possible values of $\mathrm{prod}\left(x\right)$. Since the product is taken modulo 7, the possible values of the product modulo 7 are from the set $\{0,1,2,3,4,5,6\}$.
However, let's look at the alphabet $\Sigma =\{1,2,3,4\}$. None of the symbols in $\Sigma$ is a multiple of 7, nor can any product of these symbols be a multiple of 7 because 7 is a prime number and none of the prime factors of the numbers in $\Sigma$ (which are 2 and 3) can multiply to form a multiple of 7. Thus, the modulo 7 product of any string over $\Sigma$ can never be 0.
Consequently, the only possible values of $\mathrm{prod}\left(x\right)$ for any $x\in {\Sigma }^{*}$ belong to the set $\{1,2,3,4,5,6\}$.

Let us define the states of our DFA. The state of the DFA after reading a string $x$ must keep track of the current product modulo 7. Since there are 6 possible modulo values $\{1,2,3,4,5,6\}$, we can define 6 states corresponding to each of these values:
State $q_1$: current product modulo 7 is 1 (this is also the start state because $\mathrm{prod}\left(ϵ\right)=1$).
State $q_2$: current product modulo 7 is 2 (this is the only accepting/final state).
State $q_3$: current product modulo 7 is 3.
State $q_4$: current product modulo 7 is 4.
State $q_5$: current product modulo 7 is 5.
State $q_6$: current product modulo 7 is 6.

To verify if all these 6 states are necessary and distinct (i.e., that the DFA is minimal and no states can be merged), we can check the transition behavior.
For any state $q_i$ representing a product $i��\{1,2,3,4,5,6\}$ and an input symbol $a\in \Sigma$, the transition goes to state $q_j$ where:
$j=(i\times a)\mathrm{mod}7$.

Let's check if all states are reachable from the start state $q_1$:
- From $q_1$, reading '1' keeps us in $q_1$.
- From $q_1$, reading '2' goes to $q_2$.
- From $q_1$, reading '3' goes to $q_3$.
- From $q_1$, reading '4' goes to $q_4$.
- From $q_1$, reading '2' then '3' gives $(1\times 2\times 3)\mathrm{mod}7=6$, reaching $q_6$.
- From $q_1$, reading '3' then '4' gives $(1\times 3\times 4)\mathrm{mod}7=12\mathrm{mod}7=5$, reaching $q_5$.
Thus, all 6 states are reachable.

Next, we verify that all 6 states are distinguishable (meaning no two states are equivalent):
Two states $q_i$ and $q_k$ (where $i\ne k$) are distinguishable if there exists a string $w$ such that starting from $q_i$ and reading $w$ leads to the accepting state $q_2$, while starting from $q_k$ and reading $w$ leads to a non-accepting state, or vice versa.
Mathematically, we want to find a string $w$ such that:
$(i\times \mathrm{prod}(w\left)\right)\mathrm{mod}7=2$ but $(k\times \mathrm{prod}(w\left)\right)\mathrm{mod}7\ne 2$.
Since 7 is a prime number, the integers modulo 7 form a field ${\mathbb{Z}}_7$, meaning every non-zero element $i\in \{1,2,3,4,5,6\}$ has a unique multiplicative inverse $i^{-1}\mathrm{mod}7$.
Thus, the equation $(i\times P)\mathrm{mod}7=2$ has a unique solution $P=(2\times i^{-1})\mathrm{mod}7$.
Since $i\ne k$, the value of $P$ required to reach the accepting state from $q_i$ will be different from the value required from $q_k$.
Since we can generate any modulo value in $\{1,2,3,4,5,6\}$ using strings over $\Sigma$, every pair of states is distinguishable. Therefore, no states can be merged.

Hence, the minimum state DFA requires exactly 6 states.