Short or long division?

$\LARGE{ \begin{array}{rll} \phantom{0}\ \boxed{{\phantom0}y} && \\[-2pt] \boxed{{\phantom0}x}\ \enclose{longdiv}{\boxed{\phantom0a} \ \boxed{\phantom{0}z}}\kern-.2ex \\[-2pt] \underline{\boxed{\phantom0b} \ \boxed{7}} && \\[-2pt] \boxed{{7}} \end{array} }$ It can solved with various ways however simple way to solve this problem is to focus the last digit obtained from the product of $x$ and $y$ . $\begin{aligned} x\times y = b7 = 10b+7 \end{aligned}$ The last digit is $7$ which is an odd integer which can lead to conclusion that $x$ and $y$ neither consecutive integers nor even integers Since the product of two consecutive integers and even is always even. So we are just left with the odd digits $1, 3,5,7,9$ as we can reject the even digits.

Possible cases are $\begin{aligned} &x\times y = 7\times 1= 7 \\& x\times y = 9\times 3 = 27 \\& z = 4 \end{aligned}$ Also $\begin{aligned}& 10a + z -(10b+7)= 7 \\ & \phantom{ccc} a-b = 1 \end{aligned}$ If $x\times y = 7$ , then $a =1$ and $b=0$ however the remainder will be $0$ which becomes false . So only possible product of $x\times y =27$ then $a =3$ and $b=2$ . The sum therefore is $21$ .