10
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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:

Argon is about 1%?
One-half of twelve divided by zero equals zero?
Repeat repeat lost lost primarily primarily with with skirts skirts?

Accessible version:

         ???
     -------
????|???????
     ITSSE
     -----
       ERLT
       LOVE
       ----
       IAMVE
       IEITS
       -----
        LTAE
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2
  • $\begingroup$ I think I solved the 1st and 3rd cryptic clues, but I'm massively stuck on the second one... $\endgroup$ Jul 21, 2023 at 1:40
  • $\begingroup$ Meanwhile I can only get the second one 😅 $\endgroup$
    – samm82
    Jul 21, 2023 at 2:58

2 Answers 2

7
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The cryptic clues:

AIR = AR + I [&lit, since air is about 1% argon)]
LOVE ["zero"] = (-twe)LVE + O
ITERATE ["repeat"] = RE(-p "lost primarily primarily")EAT* ("repeat lost") next to ("with") (-w)IT(-h) ("with skirts [as in avoids] skirts [as in edges]") [Thanks to OP for helping me clean this up!]

The alphametic problem:

Alphametic problem

Some easy steps:

The first thing to notice is that I times LOVE is LOVE, making I=1. Now look at the first subtraction. In the hundreds place, we must have E > S since the minuend (1TERA) must be greater than the subtrahend (1TSSE). But since E-S = E, this forces S = 0. Finally, in the first product, we have A times LOVE = 1T00E. Progress thus far:

First steps

The next digit:

In the last subtraction, we see 1AMVE is the minuend and 1E1TO is the subtrahend, so we must have A > E. However, in the first subtraction, 1TERA - 1TSSE = ERL. Looking at the ones digit, since A > E, we must have A-E = L with no borrowing. Moreover, in the second subtraction the minuend ERLT must be greater than the subtrahend LOVE, forcing E > L. Hence we have A > E > L > 1, the last inequality since 0 and 1 are already assigned.

But now look at the first product A times LOVE. It is a 5-digit number with the ten-thousands digit 1. If L at least 3, then E is at least 4, forcing A to be at least 7: in this case the product of A times LOVE would be at least 21,000. Thus we must have L = 2. Progress thus far:

Next digit

Next steps:

Notice that in all three LOVE products, the final digit is either E or 0. Looking at LOVE times R (the final product), we see the ones digit of the product is 0, which means one of E/R is 5, and the other is even. Now look at the first product A times LOVE; if E were 5, then A would be 7 and the tens digit of the product would be 7 × V + 3 (mod 10), which would force V to be 1, but 1 is already assigned. Hence R must be 5, and E is even. As A = E+2, we must have A=8 and E=6, or A=6 and E=4. But in the second subtraction, in the thousands place we have an E in the minuend and an L in the subtrahend, which forces E to be at most L+2, hence E = 4 and A = 6. Progress thus far:

Next steps

Finishing up:

Now in the last subtraction, we have V-T = 6, as there is no borrowing in the ones. If V > T, then the only remaining pair meeting that criterion is V = 9, T = 3, but in the second subtraction, in the tens digit, that would force 2 - V to yield either a 2 or 3 digit as M, but those two values would already be assigned. Thus V < T and 10+V-T = 6, or T-V = 4. The only remaining assignment allowing that difference is T = 7 and V = 3. This forces M = 9, and thus we have the answer:

Final answer

The final solution is:

SILVER ATOM

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7
  • $\begingroup$ you've just about got it! you might want to check your letter counts in clue #3 - you've got the correct answer but the work isn't quite right $\endgroup$
    – juicifer
    Jul 21, 2023 at 20:09
  • $\begingroup$ @juicifer Ah, yep...I see what I did. Let me try to repair. $\endgroup$ Jul 21, 2023 at 20:13
  • $\begingroup$ the intended solution was {"jvgu" (arkg gb) JVGU "fxvegf" (nibvqf) "fxvegf" (rqtrf)}, but you got most of the way there so I'll go ahead and accept it. nice solve! $\endgroup$
    – juicifer
    Jul 21, 2023 at 20:29
  • $\begingroup$ @juicifer Ah, I see...that's much cleaner. I'll edit the intended into the answer. $\endgroup$ Jul 21, 2023 at 20:30
  • $\begingroup$ Looks like I need to be faster with my write ups. :-/ $\endgroup$
    – fljx
    Jul 21, 2023 at 21:01
1
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First, the clues:

Argon is about 1%?

Argon makes up just under 1% of our atmosphere. The word we want here is AIR.

One-half of twelve divided by zero equals zero?

One half of "twelve" is "lve" with a zero (or O) inserted gives us LOVE, which is also a score of zero in tennis.

Repeat repeat lost lost primarily primarily with with skirts skirts?

Not sure on the wordplay for this one, but the answer is ITERATE. (I worked backwards from the alphametic to get this one, after spotting that S must be 0).

Now for the division:

With the answers filled in we have:

          AIR
      -------
 LOVE|ITERATE
      ITSSE
      -----
        ERLT
        LOVE
        ----
        IAMVE
        IEITS
        -----
         LTAE
 

Starting with the final subtraction (IAMVE - IEITS = LTAE):
From the units we have E-S = E, so S must be 0.

And from the second multiplication (I * LOVE = LOVE) we have I = 1.
Then from the first subtraction (1TERA - 1T00E = ERL), there is no borrow from the tens/hundreds, so A-E = L. (Equivalently E+L = A).
And with 0 and 1 taken, E+L is at least 2+3. So A >= 5.

          A1R
      -------
 LOVE|1TERATE
      1T00E
      -----
        ERLT
        LOVE
        ----
        1AMVE
        1E1T0
        -----
         LTAE
 

From the first multiplication (A * LOVE = 1T00E), in the units we have A * E = E (mod 10). So either:
A is odd and E = 5. (So A = 7 or 9, and L = 2 or 4.)
Or A = 6 and E is even. (So E = 2 or 4, and L = 4 or 2.)

Continuing with the first multiplication, we now have A >= 6 and the product starts with a 1. So L is at most 3.
So L = 2.
(And either E=4, A=6 or E=5, A=7.)

From the second subtraction (ERLT - 2OVE = 1AMV), in the thousands we have E - 2 - borrow = 1.
So E = 4, and A = 6.

          61R
      -------
 2OV4|1T4R6T4
      1T004
      -----
        4R2T
        2OV4
        ----
        16MV4
        141T0
        -----
         2T64
 

Now, look at the final subtraction (16MV4 - 141T0 = 2T64) again.
There are no borrows from the known columns. If there is no borrow from the hundreds, we have V-T = 6 and M-1 = T. But with 0,1,2,4,6 already taken, there are no suitable numbers left.
So there must be a borrow, so V+10 - T = 6 and M - 2 = T. Or T = V+4 and M = T+2.
From the available digits, the only option is V = 3, T = 7 and M = 9.

Finally from the second subtraction, we have 4R27 - 2O34 = 1693.
So R = 5 and O = 8

          615
      -------
 2O34|1745674
      17004
      -----
        4527
        2834
        ----
        16934
        14170
        -----
         2764
 

And finally, the word/phrase we get by putting the letters in order is:

SILVER ATOM

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