According to the last condition the remainder of this division $\overline{aaaa}$ by ${\overline{bb}}$ is $\overline{(a-b)(a-b)} = 11(a-b)$. This combined with the fact that ${\textstyle \frac{\overline{aaaa}}{\overline{bb}} = \frac{1111a}{11b} = \frac{101a}{b}}$, yields $101a = qb + a-b$, i.e. ${\textstyle q - 1 = \frac{100a}{b}}$, implying $(1) \;\; b|100a$. Also we have ${\textstyle 140 < \frac{101a}{b} < 160}$, or alternatively, $(2) \;\; {\textstyle \frac{140}{101}b < a < \frac{160}{101}b}$. Clearly ${\textstyle \frac{140}{101}b < a \leq 9}$, which means ${\textstyle b < \frac{909}{140} < 7}$. Moreover, if $b=1$, then ${\textstyle 1 < \frac{140}{101} < a < \frac{160}{101} < 2}$. In other words, $2 \leq b \leq 6$. If $b \in \{2,4,5\}$, then (1) is satisfied. If $b \in \{3,6\}$, then (1) is satisfied iff $3|a$. Using this fact combined with (2), the result is: $b=2 \;\; \Rightarrow \;\; {\textstyle 2\frac{78}{101} < a < 3\frac{17}{101}} \;\; \Rightarrow \;\; a=3$. $b=3 \;\; \Rightarrow \;\; {\textstyle 4\frac{16}{101} < a < 4\frac{76}{101}}$. $b=4 \;\; \Rightarrow \;\; {\textstyle 5\frac{55}{101} < a < 6\frac{34}{101}} \;\; \Rightarrow \;\; a=6$. $b=5 \;\; \Rightarrow \;\; {\textstyle 6\frac{94}{101} < a < 7\frac{93}{101}} \;\; \Rightarrow \;\; a=7$. $b=6 \;\; \Rightarrow \;\; {\textstyle 8\frac{32}{101} < a < 9\frac{51}{101}} \;\; \Rightarrow \;\; a=9$. Summa summarum, there are four pair $(a,b)$ of digits satisfying the given conditions, namely $(a,b) = (3,2), \: (6,4), \: (7,5) \: (9,6)$.