# Cryptic Division 12: And All the Men and Women Merely Players

This is a word division puzzle which uses cryptic clues. If you're unfamiliar with either or both of those, you can click the associated link.

In order to solve the alphametic, you'll first need to fill in the dividend, divisor, and quotient by solving the cryptic clues. I've left the enumerations off to provide a bit of extra challenge. Once you fill those in, the puzzle should be solvable with only arithmetic and logic. The solution is a 10-letter word or phrase found by ordering the letters from 0 through 9. A complete answer should provide this solution along with explanations of the cryptic clues and your path through the alphametic.

As always, I've created an interactive version that will autofill from the grid to the clues and vice versa. Have fun!

Clues:

• Competitors at heart, on Parisian river, (mostly) sorted alphabetically, going from east to west? This is most peculiar
• Place to hold water sports in Nanterre initially given hasty approval
• It makes return in table-tennis

Accessible version:

         ???
-------
????|???????
ZEZAA
-----
ITSIS
IBTIB
-----
KNUKT
KSZIA
-----
BAAN

• Looks like link goes to old puzzle... Commented Jul 31 at 20:44
• @JeremyDover oops! thanks for the heads up! should be the right one now Commented Jul 31 at 20:53

The clues:

ZANIEST - (_T_ + SEIN(-e) + AZ)<
SINK - S_ I_ N_ + K (hasty approval)
BAT - (TAB<)le-tennis

The alphametic:

From comparing the dividend ZANIEST with ZEZAA, we find that A > E, but in the first subtraction we have E-A in the ones digit, so we know that a borrow must be involved. Thus 10+E-A = I. However, from the thousands digit in this first subtraction, we see that either A-E = I or (A-1)-E = I, the latter case occuring if there was a borrow. But this latter case would force 10+E-A = (A-E)-1, or 11 = 2(A-E), which cannot occur by parity. So there is no borrow from the thousands digit, and we have A-E = 5 = I. This also shows A > 5 and E <= 4.

From the 10s digit in the first subtraction, we know there was a borrow for the ones digit, and A > 5, so this tells us that 14-A = S. This tells us that S > 5 as well. Finally from the hundreds digit, we know there was no borrow from the thousands, but there was a borrow for the 10s, so we have N-1-Z = T.
Look at the products of each of B, A and T with SINK. notice that in the first product B×SINK = T×SINK modulo 10, whence (B-T)×K = 0 mod 10. K is not 0 since these values are not K, nor is B-T, hence one (or both) of these values must be divisible by 5, and one or both must be divisible by 2. K is neither 0 nor 5, so 5 must divide B-T, and since |B-T| < 10, we must have |B-T| = 5 and K is even. But now look at the thousands digit of the second subtraction. We know |T-B| = 5, and we cannot borrow, so the hundreds must have borrowed, giving us T-1-B = K, which shows that T-B = 5, and K = 4.

Now look at the last subtraction. In the thousands digit, we have N > S > 5. Hence, we know that A, N, S and T are the four digits greater than 5. In the tens digit, there is definitely a borrow. The only question is if there was a borrow in the ones as well. But regardless, it shows that A is either 8 or 9. As 14-A = S, and S > 5, this forces A = 8 and S = 6. Now in the ones digit, if T were 9, then N would be 1, but we know N > 5. Hence T is 7 and N is 9.

Finishing up, the second subtraction ones digits gives us B = 2, and the 10s digits gives us U = 0. The first multiplication resolves the remaining digits, and the final answer is UZBEKISTAN!

• Nice! You beat me to it. I got tripped over my math because I was solving this backwards (trying to get the letter values) but got pretty close. Commented Jul 31 at 23:18