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Let $L = \{\text{madeeasy2016}\}$ over $\Sigma = \{m,a,d,e,s,y,2,0,1,6\},$ $L_1=\text{prefix}(L)$ and $L_2= L_1 \Big /{} { \Sigma ^*}$. The number of strings in $L_2$ (assume $L_1$ and $L_2$ do not include empty string) are ______.
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$L =\{\text{madeeasy2016}\}$

$L_1 = \text{prefix}(L)$, excluding empty string

$L_1= \{m,ma,mad,made,madee,madeea,madeeas,madeeasy,\\ \qquad \quad madeeasy2,madeeasy20,madeeasy201,madeeasy2016\}$


In general Right Quotient of $L_1$ with $L_2$ means $L_1\Big /L_2=\{x\;:\;xy\in L_1 \;\text{for some}\;y \in L_2\}$

so In our problem

$L_2=L_1\Big /\Sigma^*=\{x\;:\;xy\in L_1 \;\text{for some}\;y \in \Sigma^*\}$

for $y=\epsilon$, $L_2=L_1 \Big / \Sigma^* = L_1$

Number of strings in $L_2 =$ Number of strings in $L_1  =12$

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Lets explore the quotient operator:

L1/L2 = ?

? is every 'x' that satisfies x.y L1 where y ∈ L2

Now, consider any arbitrary language L. What is L/∑* ? 

So, take...      x.y ∈ L     as 'y' can range from epsilon to every string possible, you will find that x is actually a prefix for L. How? Say y='b', and L contains some string like aabbb. So x becomes 'aabb'. So, if you consider all possible x (which is nothing but the quotient) , you will get prefix(L).

Now, coming to your question, L1/∑* is nothing but prefix(L1). Thus, L2=Prefix(L1) which inturn is prefix for given string. So, just calculate prefixes, that's it :) 

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