Cryptogram 22

The long multiplication can be written as:

$\begin{aligned} (10A+B) \times 7 & = 400 + 10C + C \\ 7(10A+B) & = 400 + 11C \end{aligned}$

Since both sides of the equation are integers, the RHS is divisible by 7, that is:

$\begin{aligned} 400 + 11C & \equiv 0 \text{ (mod 7)} \\ 1 + 4C & \equiv 0 \text{ (mod 7)} \\ \implies C & = 5 \end{aligned}$

Then $7 \times \overline{AB} = 455$ $\implies \overline{AB} = \dfrac {455}7 = 65$ $\implies A = 6$ , $B=5$ , and $A-B = \boxed 1$ .