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This is a word division puzzle which uses cryp- actually, this one uses printer's devilry. 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 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:

Wayne Gretzky was a stare DM onto 'er, not a player for a football club in Vancouver
University's Sting lineup includes a player for a football club in Darwin
Goaltender Zac Barand, some former player for a football club in Luton

As mentioned, these are printer's devilry clues, and I've included an additional definition at either the start or end of the clue.

Accessible version:

        ???
     ------
????|??????
     TEPR
     ----
      IENER
      ILTLT
      -----
       RTRL

Hint/Disclaimer (posted 2 April):

The definition for clue #2 (the Darwin one) is a bit obscure but I wanted to fit with the theme I'd established with the other two clues. That said, once you figure out the "twist" to this puzzle, you should be able to confirm that the answer fits the definition with a quick search on the engine of your choosing.

Hint 2 (posted 3 April):

Some significant progress has been made on the first two clues, so I suspect clue #3 (the Luton one) will be the one to turn things around, as it were. To that end: the clue (including the answer) involves a real person and real facts about that person - as well as one word which is an opinion I hold about him and holds no bearing on any of the other facts about him.

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  • 3
    $\begingroup$ After doing a lot of googling and giving up, my phone is now sending me Luton Football scores every 20 minutes. Puzzles have consequences! $\endgroup$ Commented Apr 6 at 21:27

3 Answers 3

3
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OK I think I've got it now.

Wayne Gretzky was a star Edmonton Oiler, not a player for a football club in Vancouver. NOIL, which reversed is LION (the mascot of a football club in Vancouver).

University's Starting lineup includes a player for a football club in Darwin. TAR, which reversed is RAT (which presumably has something to do with a football club in Darwin??)

Goaltender Zac Barretta, handsome former player for a football club in Luton. RETTAH, which reversed is HATTER (the name of a football club in Luton).

And so the alphametic becomes this:

enter image description here

which can be solved like this:

enter image description here

For a full explanation of how to solve the alphametic, see ACB's answer below.

Putting the letters in order from 0-9 means the final phrase is:

APRIL ONETH

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  • $\begingroup$ this is the correct answer, nice job! could you add some details on your solve path for the alphametic? once you do, I'll give you the checkmark $\endgroup$
    – juicifer
    Commented Apr 4 at 2:13
  • $\begingroup$ for the darwin clue: uggcf://jjj.snprobbx.pbz/HavEngfQnejva/ $\endgroup$
    – juicifer
    Commented Apr 4 at 2:14
  • 1
    $\begingroup$ oh, and I've also just realized you didn't include the solution phrase! that's pretty important too haha $\endgroup$
    – juicifer
    Commented Apr 4 at 4:31
  • 1
    $\begingroup$ still waiting on a solve path. are you willing to edit that into your answer? $\endgroup$
    – juicifer
    Commented Apr 5 at 12:27
2
+100
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Solve path for the alphametic

First let's build some inequalities.

Consider

        RAT
     ------
LION|HATTER
     TEPR
     ----
      IENER
      ILTLT
      -----
       RTRL

Since both E and R are brought down from the dividend, IENE must be less than LION (divisor). Therefore I<L. The former also confirms that A=0.

        R0T       0-A
     ------       1-
LION|H0TTER       2-
     TEPR         3-
     ----         4-
      IENER       5-
      ILTLT       6-
      -----       7-
       RTRL       8-
                  9-

In addition, RTRL(remainder)<LION(divisor) implies that R<L.

   H0TT
  -TEPR
  _____   implies T+1=H
    IEN

Or in other words,

T and H are adjacent numbers.

        R0T       0-A
     ------       1-
LION|H0TTER       2-
     TEPR         3-
     ----         4-
      IENER       5-
      ILTLT       6-
      -----       7-
       RTRL       8-
                  9-

Since LION×R=TEPR, R≠1 and also L<T.

So far we have,

(I,R)<L<T<H

Now look at

        R0T       0-A
     ------       1-
LION|H0TTER       2-
     TEPR         3-
     ----         4-
      IENER       5-
      ILTLT       6-
      -----       7-
       RTRL       8-
                  9-

Or simply,

 ENER
-LTLT
 ____
 RTRL

This directly implies that L<E. Other than that, can you spot the symmetry here?

 EN ER
-LT LT
 _____
 RT RL

It is previously deduced that R<T. Thus we will have to borrow from E in order to subtract T from R. Therefore

10+R-T=L (a)
(E-1)-L=R

From the symmetry argument, we get

10+N-T=T (b) or
10+N=2T

Therefore N is even and as N>0, T≥6.
Also since there is at least one number greater than T, T≤8. Therefore N≤6. Thus
N=2/4/6
T=6/7/8

Likewise from (a) and (b),

L<T⇒R<N (for future reference)

Now

        R0T       0-A
     ------       1-
LION|H0TTER       2-
     TEPR         3-
     ----         4-
      IENER       5-
      ILTLT       6-
      -----       7-
       RTRL       8-
                  9-

T-R=N.
Also we have 10+R-T=L.
Combining the two,

L+N=10.
So L is also even. But there are at least 3 numbers below L. Therefore L≥4.
Hence (L,N)=(4,6).

We previously deduced that R is less than both L and N.

So R=2 or 3.

Now we go back to the multiplication

   LION
  ×   R   where N=4 or 6
  _____         R=2 or 3
   TEPR

This makes it obvious that

N=6 and R=2.
Thus L=4.

        20T       0-A
     ------       1-
4IO6|H0TTE2       2-R
     TEP2         3-
     ----         4-L
      IE6E2       5-
      I4T4T       6-N
      -----       7-
       2T24       8-
                  9-

Then the remaining letters (T, H, E, P, I, O) can be easily deduced.

        208       0-A
     ------       1-P
4I56|908872       2-R
     8712         3-I
     ----         4-L
      37672       5-O
      34848       6-N
      -----       7-E
       2824       8-T
                  9-H

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  • $\begingroup$ I shouldn't be that surprised, but this is a completely different path from the one I was expecting! though obviously no less valid :) my in was that rot13(A * E = E naq A * G = G, fb A zhfg or rvgure 1 be 6; ohg vs A=1, gura jr unir G-E=1, nf jryy nf E-G=Y, zrnavat gung Y=9, ohg Y pna'g or 9 orpnhfr n) gur qvivfbe pnaabg gura zhygvcyl gb nabgure 4-qvtvg ahzore naq o) gur obggbz fhogenpgvba jbhyq pnhfr n obeebj sebz V-V=0. gurersber A=6, Y=4, naq E naq G ner lbhe bgure gjb rira ahzoref), at which point everything falls apart $\endgroup$
    – juicifer
    Commented Apr 10 at 13:00
  • $\begingroup$ That is significantly easier than what I did! $\endgroup$
    – ACB
    Commented Apr 10 at 13:23
0
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Some starters:

Wayne Gretzkey was a star edmontoN ... er, not a player for a football club in Vancouver
University's stARTing lineup includes a player for a football club in Darwin (or TAR)

Unfortunately, I don't know enough about hockey or football to full answer for 1, or distinguish the two choices for 2.

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2
  • $\begingroup$ I got the same as you and in addition rot13(ABVY jbhyq frrz gb or gur nafjre sbe gur svefg pyhr nf Tergmxl jnf n fgne Rqzbagba Bvyre. Hasbeghangryl nffhzvat gur nafjre gb gur frpbaq pyhr vf rvgure NEG be GNE, V jnf hanoyr gb svaq n crezhgngvba bs qvtvgf gung jbhyq svg jvgu jung jr'er tvira. Fb rvgure V'ir znqr n zvfgnxr fbzrjurer be bar bs gubfr nafjref vf vapbeerpg.) $\endgroup$
    – SQLnoob
    Commented Apr 1 at 23:00
  • $\begingroup$ rot13(Gur OP Yvbaf ner gur PSY grnz va inapbhire, juvpu vf ABVY va erirefr.) which confirms @SQLnoob $\endgroup$ Commented Apr 2 at 12:49

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