32 A four digit number consists of two distinct pairs of repeated digits (for example 2211, 2626 and 7007). Find the total number of such possible numbers that are divisible by 7 or 101 but not both.
32 Answer: (97)
We let the 2 distinct digits be $A$ and $B$ with $1 \leq A \leq 9,0 \leq B \leq 9$ and $A \neq B$.
Case 1: $A B A B=101 \times A B$. There are $9 \times 9=81$ possibilities.
However $A B$ must not be a multiple of 7 ( 12 possibilities excluding 07 and 77 ) $\Rightarrow$ There are $81-12=69$.
Case 2: $A A B B=11 \times(100 A+B)$. There are 11 possibilities that are factors of 7 . $(A, B)=(1,5),(2,3),(3,1),(3,8),(4,6),(5,4),(6,2),(6,9),(7,0),(8,5),(9,3)$.
Case 3: $A B B A=11 \times(91 A+10 B)$. $B$ must be 0 or 7 . There are 17 possibilities.
$\therefore$ Total such possible number $=69+11+17=97$.