11
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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:

Is 9 plus 10 divided by 5 ultimately a square number?
Ceiling of discontinuous function given by 20 - y (didn't show work!)
Limit found in Serre's geometry paper? The limit does not exist!

Accessible version:

         ???
     -------
????|???????
     SUIIU
     -----
      HANTE
      XNHSN
      -----
       STEHN
       SUIIU
        ----
        XHSU
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2
  • $\begingroup$ There are only 9 letters. $\endgroup$ Sep 25, 2023 at 23:04
  • $\begingroup$ @WeatherVane I suppose it's possible that the blanks introduce another letter. However, I've solved the first clue and it doesn't introduce another letter. Also I really hope I don't have to read this paper $\endgroup$ Sep 25, 2023 at 23:13

1 Answer 1

11
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Cryptic clues:

We get a bit of a hint from (1) the fact that the last two letters of the dividend have to be EN since those have been "copied down" in the long-division process, and (2) the two occurrences of SUIIU, indicating that the quotient has matching first and third letters. Oh, and (3) nine letters are given initially, and the divisor and dividend don't yield any new ones, so if we're supposed to be finding all ten then probably the quotient will contain a new letter.

Is 9 plus 10 divided by 5 ultimately a square number?

'S can be an abbreviated form of IS. IX is 9. TEN is 10. E is "fivE ultimately". So we have S IX TEN "divided by" (i.e., split up by inserting) E, getting SIXTEEN, which is indeed a square number.

Ceiling of discontinuous function given by 20 - y (didn't show work!)

20 is TWENTY from which we can remove Y to get TWENT and then W (work) to get TENT. And a tent is used as a ceiling for something like a circus, a function (in the sense of event) that happens sporadically (hence discontinuous).

Limit found in Serre's geometry paper? The limit does not exist!

This clue actually requires some knowledge about mathematics (though it can be provided by Google). One of J-P Serre's most important papers, in the field of algebraic geometry, is usually referred to by the initials of its (French) title as GAGA. Remove its last letter ("the limit does not exist") and you get GAG, which does indeed kinda mean a limit (or to limit).

So now we have

                  G A G
        _______________
T E N T ) S I X T E E N
          S U I I U
          ---------
            H A N T E
            X N H S N
            ---------
              S T E H N
              S U I I U
              ---------
                X H S U

from which we can see that

H=X+1 and A<N (from the left end of HANTE-XNHSN=STEH), so our final answer has ...XH... in it somewhere. (We are presumably supposed to solve the alphametic "honestly", but I cannot promise not to let my solution be guided by thoughts about the possible final answer.)

A few other easy remarks:

S<T because otherwise we would get a longer quotient. Modulo 10, N=2U (from the last digit of the last subtraction). None of G, A, T, S, H, X can be zero (because of leading zeros or zero products). So 0 is one of EIN. Also, A and G are both > 1 since the products they produce are longer than TENT is.

Now, let's be a little fancier.

We have G x TENT = SUIIU so mod 1001 we have G x EN0 = S0II0 or G x EN = S0II. Clearly G x EN < 1001, so we must more specifically have G x EN + S00S = S0II, or G x EN + S = II. So, handwavily, G and E are fairly small and I is fairly big. [Don't you mean "I am fairly big?" -- Ed.] We'll try to get more from this later, but for now let's just note that in particular it implies that I is another letter that can't be 0; so 0 is either E or N. Oh, and it can't be N either because then G x EN + S = II would imply that S = I. So E is 0.

At this point (in fact, before spotting the last inference there)

I confess that I looked at the remaining possibilities for where E,X,H,N,U can go -- there aren't all that many -- and spotted that the final answer is almost certainly going to be EXHAUSTING. But let's try to proceed honestly, as promised.

Well,

we have some more mod-1001 hijinks to do, because we can play the same game with the other product we know and get A x EN = X0HS mod 1001 and hence, again, A x EN + X = HS. And we also just found that E=0, so this means A x N + X = HS. And of course our earlier equation is now G x N + S = II.

And

subtracting these from G x TENT = SUIIU and A x TENT = XNHSN and dividing by 1001 gives G x T = SU and A x T = XN.

Next,

at the right end of the first subtraction we see that mod 10 E-U=T, and we know E=0 so this tells us that U+T=10; similarly, in the second subtraction we see that mod 10 E-N=H, so N+H=10. And if GxT=SU then mod 10 we now know that GxT = -T, or (G+1)xT=0. So we might have G=9, or G=4 and T even, or G odd and T=5. The last of these won't do, though, because U+T=10 and U,T can't be equal. So either G=9, or G=4 and T is even. In the latter case T can't be 0, because that's E, and it can't be 4 or 6, because then T or U would equal G. So either G=9, or G=4 and T is 2 or 8. Further, we can't have G=4 and T=2 because we need GxT = SU and S isn't 0. So either G=9, or G=4 and T=8 (and U=2); but this last possibility is also impossible because N = 2U mod 10, and if U=2 then this means N=4, colliding with G. So, finally, we have found that G=9.

We now know

9 x N = II - S; 9 x T = SU.
Thinking about what multiplication by 9 does to single digits, the second of these means that S+U=9 (and since T+U=10 this means T=S+1). But wait, we also know that N = 2U or N = 2U-10. So II = 9 x N + S = 9 x (2U[-10]) + 9-U = 17U + (9 or -81). Running through single-digit multiples of 17, we can get a multiple of 11 this way either with U=4 or with U=8. The former gives N=8, S=5, T=6. The latter gives N=6, S=1, T=2.

Two possibilities, but probably we can show that one of them is impossible

and our guess at the final answer makes it clear which one we should try to refute. Suppose, then, that we have (E=0, G=9,) U=8, N=6, S=1, T=2. Well, we also have A x T = XN; that is, A x 2 = X6, but this is no good because the LHS is obviously at most 18 but X can't be 0 or 1 since those are already taken. So instead we have (E=0, G=9,) U=4, N=8, S=5, T=6.

Now once again

look at the relations involving A. We have A x 6 = X8, which means A is 3 or 8; the latter collides with N, so A=3 and then X=1. And we have 3 x 8 = H5 - 1 so H=2. And now the only unallocated letter is I and the only unallocated digit is 7, and we're done.

The completed division is:

                  9 3 9
        _______________
6 0 8 6 ) 5 7 1 6 0 0 8
          5 4 7 7 4
          ---------
            2 3 8 6 0
            1 8 2 5 8
            ---------
              5 6 0 2 8
              5 4 7 7 4
              ---------
                1 2 5 4

and the final answer is

EXHAUSTING.

Credit where due: I mis-parsed the first cryptic clue a bit and entirely unjustly made unhappy noises about it as a result; Jeremy Dover, in comments, pointed out the error of my ways.

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  • 5
    $\begingroup$ I think the E in the first clue is fivE, not since E is the fifth letter of the alphabet. $\endgroup$ Sep 26, 2023 at 2:14
  • 1
    $\begingroup$ nice work! as usual I had a completely different solve path to get this one, and it was definitely cool to see mod 1001 come out of the woodwork :D. @JeremyDover is spot on as to my intentions for that first clue btw, but otherwise 100% correct $\endgroup$
    – juicifer
    Sep 26, 2023 at 2:43
  • 1
    $\begingroup$ @JeremyDover oh, thank goodness for that. Will fix my answer. [EDITED to add:] Now done. Thanks. $\endgroup$
    – Gareth McCaughan
    Sep 26, 2023 at 15:53

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