Let ? be the divisor. Consider the multiples of ?:

dw (k*?)

pq (d*?)

vwk (u*?)

vx (v*?)

us (p*?)

rk (q*?)

The one that caught my attention was vx. The unknown divisor multiplied by v was vx, which meant that the divisor had to be less than 20. And also, since x-s=x, s equals 0. And at most, v is 4, but at least 1.

Since k=k, u was 5 more than q. Also, it meant that ? was even, namely 12, 14, 16 or 18. (Why not numbers underneath? Because that means the dividend would have to be smaller.) x is either 2, 4, 6 or 8.

Since ? was under 20, that meant v was 1, from vwk and rk. The divisor was vx. k + u = 11. And since ? is an even number, k must also be an even number, which means that u is an odd number, which rules out 12 and 16 as ?, and that means x is either 4 or 8.

Two things to consider: k + u = 11, u – q = 5; which means that u, an odd number, is larger than 6, which means that it can only be 7 or 9. In either of those situations, p is 5. If u = 7, k = 4, and if u = 9, k = 2.

Consider x = 4. The divisor is 14, and pq would be 56. Then d would also be 4, and this would be false. So therefore x = 8, and d = 3, q = 4, u = 9, k = 2, r = 7, w = 6.

Therefore:

s = 0

v = 1

k = 2

d = 3

q = 4

p = 5

w = 6

r = 7

x = 8

u = 9

And the final solution of rewriting the division with the digits and the dividend is:

That’s it.

Side note: I’m going to delay the post on normal numbers. 😐

Thank you to the many people at https://enigmaticcode.wordpress.com/ that spent their time uploading Enigma puzzles. I appreciate it. This stuff is fun. Thank you.