# Cryptic Division 11: At the Water's Edge

Somewhere in an alternate universe...

Dear Puzzling,

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!

Today, I arrived at my final stop on a four-leg trip around the globe. I've heard some people take trips like these with over 60 destinations! That seems like too much planning, if you ask me. Anyway, I've provided some notes about the things I've seen on this trip below. As you can see, I've spent a lot of time near the water, and that's not about to change now. I'll be enjoying my time on the banks of one of the world's most famous rivers, surrounded by some of the oldest architectural feats in the world, and also some of the newest. Can you guess where I am?

Clues Voyage Notes:

Aquatic creature seen off the coast of Washington, Oregon, and California
It's held in water - e.g. at Tangier
Gladys is visiting Philippine port on Pacific coast with member of Gladys's support group?

Accessible version:

         ???
-------
????|???????
ROOCP
-----
AIAAT
PGCI
-----
REAYA
ROOCP
-----
YGR


Many thanks to the incomparable Jafe for letting me add a little bit of non-canon Gladys flavor to this puzzle!

• Gladys and the Case of the Substitute Ghostwriter Commented May 30 at 14:12

First cryptic clue

is OR + CA making ORCA, which must be the divisor.

Second cryptic clue

is a hidden-substring one with solution REGATTA, which must be the dividend.

We have

a full set of 10 different letters -- ACEGIOPRTY -- so our third word must be made only from those.

Third cryptic clue

has first and last letters the same; presumably the first "Gladys" means either G or I; so far as I know there are no two-letter ports in the Philippines, but maybe there are some with two-letter abbreviations. Most promising option seems PIP; ah, there was a band called "Gladys Knight and the Pips", so PIP it is. (There's at least one Philippine port whose name could plausibly be abbreviated to PP, but others in the comments have very politely pointed out that I'm an idiot: it's the leftmost (port in nautical terminology) letter of "Philippine" plus one side (coast) of "Pacific".)

Now, let's look at the alphametic.

We have RE - RO = A in the first subtraction but RE - RO = (nothing) in the second. So clearly E=O+1, A=1, and there's a borrow in the second case but not the first. (We could also tell that A=1 from the second subtraction.) Then the very last pair of subtracted digits tells us that P+R=11; and the rightmost digits of the first subtraction tells us that either T=P+1 or T=0,P=9. And O,P,R,Y can't be 0 because they appear as leftmost digits, nor can E since E=O+1, so 0 is one of CGIT.

We do have

1-C=1 in two places, which means that either C=0 or C=9 and there's a borrow in both those places. One of those places is at the end of the first subtraction where in fact we have 1T-CP=11 (with, perhaps, a borrow propagating to the left). The only way to get a borrow from the rightmost column is to have T=0, P=9, but then we would also need C=9 and C,P can't both be 9. So there's no such borrow, and we have C=0.

So the present state of play is:

                      P I P
--------------
O R 0 1 | R E G 1 T T 1
R O O 0 P
---------
1 I 1 1 T
P G 0 I
---------
R E 1 Y 1
R O O 0 P
---------
Y G R

0 1 2 3 4 5 6 7 8 9
C A

A C E G I O P R T Y
1 0

E = O+1
T = P+1 or P-9
P+R = 11

Now,

we have P x OR01 = ROO0P. That is, P x OR = ROO. We also have I x OR01 = PG0I or I x OR = PG, so P is at least twice O. Oh, and remember that P+R=11. So, trying possible values of R from 2 to 9, we get: 9 x O2 = 2OO, 8 x O3 = 3OO, 7 x O4 = 4OO, 6 x O5 = 5OO, (5 x O6 = 6OO, 4 x O7 = 7OO, 3 x O8 = 8OO, 2 x O9 = 9OO). The parenthesized ones are obviously impossible. Each of the others has either one or two possible O to make the first digit on the right correct: 9x32=288 (no), 8x43=344 (maybe), 7x64=448 (no), 6x85=510, 6x95=570 (no, no). So in fact we have P=8, R=3, O=4. Recall that T=P+1 mod 10, so also T=9. And E=O+1, so E=5.

                      8 I 8
--------------
4 3 0 1 | 3 5 G 1 9 9 1
3 4 4 0 8
---------
1 I 1 1 9
8 G 0 I
---------
3 5 1 Y 1
3 4 4 0 8
---------
Y G 3

0 1 2 3 4 5 6 7 8 9
C A   R O E     P T

A C E G I O P R T Y
1 0 5     4 8 3 9

Nearly there.

For I x 4301 to give 8G0I we must have I=2 and hence G=6. And then the only unassigned digit is 7 and the only unassigned letter is Y, so pair those two up.

                      8 2 8
--------------
4 3 0 1 | 3 5 6 1 9 9 1
3 4 4 0 8
---------
1 2 1 1 9
8 6 0 2
---------
3 5 1 7 1
3 4 4 0 8
---------
7 6 3

0 1 2 3 4 5 6 7 8 9
C A I R O E G Y P T

A C E G I O P R T Y
1 0 5 6 2 4 8 3 9 7

And now we see that