11.3. Determine all four-digit numbers that are divisible by 11 and have the sum of their digits 1 less than the product of their digits.
Solution. The number $\overline{a b c d}$ is divisible by 11 if and only if $[a+c-(b+d)]$ is divisible by 11. Since $b+d \leq 18$ and $a+c \leq 18$, we have 2 possible cases: 1) $a+c-(b+d)=0$ and 2) $a+c-(b+d)= \pm 11$. From the condition of the problem, we have $a+b+c+d+1=a b c d$, which implies $a b c d>1$, i.e., $a, b, c, d \neq 0$.

Case 1. $a+c-(b+d)=0$.

Since $a+c=b+d$, it follows that $a b c d=2(a+c)+1$. Therefore, the product $a b c d$ is an odd number, which means all digits $a, b, c, d$ are odd numbers, leading to the double inequality $5=2(1+1)+1 \leq a b c d \leq 2(9+9)+1=37$.

It follows that at least one digit among the digits $a, b, c, d$ must be less than 3, since otherwise, we would have the false inequality $37 \geq a b c d \geq 3 \cdot 3 \cdot 3 \cdot 3=81$. Since this digit must be odd, it is equal to 1.

Let $a=1$. Then $1+c=b+d$ and $2 c+3=b c d$, which implies $b c d \leq 2 \cdot 9+3=21$. Similarly, at least one digit among the digits $b, c, d$ must be less than 3, since otherwise, we would have $21 \geq b c d \geq 3 \cdot 3 \cdot 3=27$, which is false. If $c=1$, then $2=b+d$ and $5=b d$, which implies $b=d=1$, but $5 \neq 1$. Therefore, $c \neq 1$, which means $b=1$ or $d=1$.

Let $b=1$. Then $1+c=1+d$ and $2 c+3=c d$, which implies $c=d$ and $2 c+3=c^{2}$, i.e., $c=3$ and $d=3$. Thus, we obtain the number $\overline{a b c d}=1133$. Since the initial conditions $(a+c=b+d$ and $a+b+c+d+1=a b c d)$ are symmetric with respect to $a$ and $c$ (respectively $b$ and $d$), we obtain 3 more solutions: 3113, 1331, 3311. If $d=1$, the reasoning is analogous and we obtain the same solutions. Similarly, we obtain the same solutions in the case $a \neq 1$ and one of the digits $b, c, d$ is 1.

Case 2. $a+c=b+d+11$.

We only investigate this case, as the case $a+c+11=b+d$ is analogous. It follows that $2(b+d)+12=a b c d$, which means the product $a b c d$ is an even number. We have $13=1+1+11 \leq b+d+11=a+c \leq 9+9=18$, i.e., the double inequality $13 \leq a+c \leq 18$ (*).

Since $a+c \leq 18$, then $b+d=a+c-11 \leq 9+9-11=7$. We have $a b c d=2(b+d)+12 \leq 2 \cdot 7+12=26$, which implies $a c \leq \frac{26}{b d} \leq 26$. Then $c \geq 13-a \geq 13-9=4$, which implies $a \leq 26 / c \leq 26 / 4=6.5$, i.e., $a \leq 6$. Similarly, we obtain $c \leq 6$. We get $a+c \leq 6+6=12$, which contradicts (*). Therefore, in Case 2, we have no solutions.

In conclusion, all solutions to the problem are 1133, 3113, 1331, 3311.