# The answer to Enigma #618

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.

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