1. **Stating the problem:** We are given that $A \neq 0$ and the product $ABCD \times E = 2006$. We need to find the value of $B$. 2. **Understanding the problem:** Here, $ABCD$ represents a four-digit number with digits $A$, $B$, $C$, and $D$. $E$ is a single digit number. The product of $ABCD$ and $E$ equals 2006. 3. **Expressing the problem mathematically:** Let the four-digit number be $N = 1000A + 100B + 10C + D$. Then the equation is: $$N \times E = 2006$$ 4. **Finding possible values:** Since $A \neq 0$, $N$ is between 1000 and 9999. Also, $E$ is a single digit integer (1 to 9). 5. **Check divisors of 2006:** To find $N$ and $E$, we factor 2006: $$2006 = 2 \times 17 \times 59$$ 6. **Check possible $E$ values:** Since $E$ is a single digit, possible $E$ values dividing 2006 are 1, 2, 7, 8, 9. But 7, 8, 9 do not divide 2006 evenly. Only 1 and 2 divide 2006 evenly. 7. **Try $E=2$:** $$N = \frac{2006}{2} = 1003$$ 8. **Check if $N=1003$ is a four-digit number with $A \neq 0$:** $1003$ is four-digit, and $A=1 \neq 0$. 9. **Extract digits:** $N=1003$ means $A=1$, $B=0$, $C=0$, $D=3$. 10. **Answer:** The value of $B$ is $0$. **Final answer:** $\boxed{0}$