# Cryptic Division 7: Nothing Under the Tree

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, and happy holidays!

Clues:

At first, Santa showed up loudly, which arose such a clatter
"Hauling around packages of untold regalia to everyone all over the globe is difficult!"
Promise of a gift's eventual arrival graciously accepted

Accessible version:

         ???
-------
????|???????
AIOUI
-----
IRGU
FUSS
----
OSSSS
OLAGU
-----
UOGL

• you might think this puzzle is too late for christmas, but in my opinion, we're only two days into christmas season! only 364 left to go :) Commented Dec 27, 2023 at 21:53
• Indeed, it's only the retailers who start Christmas in September. Today (27th) is the third of the twelve days of Christmas. Commented Dec 27, 2023 at 22:17

## 2 Answers

"Hauling around packages of untold regalia to everyone all over the globe is difficult!"

"Packages of" sounds like first+last letters. That would give UD RA. And "Hauling around" could be a reversal indicator. That would give ARDU. Ah, ARDUOUS. So O is the globe and I suppose "us" is everyone.

So now we have

    _____###
####)ARDUOUS

and we know

that our ten letters are AIOU DFGLRS. Looking at the alphametic we can see that R = I+1 (from the left end of the first subtraction), that D < O (same region), that F < I and O < I (left end of second subtraction), and L < S and U < S (left end of third subtraction). Also, looking at both ends of that last subtraction, we see that there is no borrow "into" the S-L=U column, so S > A and S > O. We also know U-S = S (or = S-10) so U is even. And then L = S-U = -(U-S) = -S mod 10; so S+L=10. I'm being good and not just feeding AIOUDFGLRS into an anagramming program; so far I haven't worked out what our answer is going to be yet.

OK, I can do another of the clues.

At first, Santa showed up loudly, which arose such a clatter

"At first" must be a first-letter extractor, applying at least to "Santa" and maybe more. And "loudly" must mean F. Ah, so this is FUSS: [S(anta) s(howed) u(p) F]<. (Ho ho ho, this was in the alphametic all along.)

So now we have

    _____#1#
FUSS)ARDUOUS

where we know

the 1 because of the literal FUSS in the alphametic.

And we know

that UOGL < FUSS because otherwise the last digit of the quotient is wrong, which means that U <= F, which means that U < F. Alas, we already know that F < I < R and therefore the answer can't involve RUDOLF.

Oh, duh, and of course Promise of a gift's eventual arrival graciously accepted is

IOU, substring of GRACIOUSLY. So O is 1, which definitely won't be confusing at all.

Current state of play:

    0 1 2 3 4 5 6 7 8 9
. O . . . . . . . .

I 1 U
_______________
F U S S ) A R D U 1 U S
A I 1 U I
---------
I R G U
F U S S
-------
1 S S S S
1 L A G U
---------
U 1 G L

This means that

some multiple of FUSS is at least 20130, but alas all this tells us is that F >= 2.

Ah, I think I've figured out what the answer is; it must be

DOUGLAS FIR.

I will allow my guesses to be guided by this but will of course attempt to provide a proof that these are right :-). So,

look at that first subtraction. We must have AI1 <= ARD <= AI1 + 9. That is, R = I+1 and then (10+D)-1 <= 9, or 9+D <= 9, or D <= 0. So D = 0. Looking at the next column of that subtraction, we have U-U = R. That can't happen without a borrow because D, not R, is 0. So there's a borrow from this column and R=9. And we remember that R = I+1, so I=8.

So now things look like this:

    0 1 2 3 4 5 6 7 8 9
D O . . . . . . I R

8 1 U
_______________
F U S S ) A 9 0 U 1 U S
A 8 1 U 8
---------
8 9 G U
F U S S
-------
1 S S S S
1 L A G U
---------
U 1 G L

Next,

at the right-hand end of that first subtraction we have 1-8 = G (with a borrow), so G = 3. And now look at the right-hand end of the next subtraction where we have U-S=S (without a borrow) and then 3-S=S (clearly with a borrow) so we must have U=2.

This leaves us with this:

    0 1 2 3 4 5 6 7 8 9
D O U G . . . . I R

8 1 2
_______________
F 2 S S ) A 9 0 2 1 2 S
A 8 1 2 8
---------
8 9 3 2
F 2 S S
-------
1 S S S S
1 L A 3 2
---------
2 1 3 L

and now it's just mopping up.

We just saw that 2-S yields S-with-a-borrow; that is, 2-S=S-10 or S=6. The rightmost digit of the quotient is 2, so 1LA32 = 2xF266, so A=5. Then in the last subtraction we have 6-L=2 with no borrowing, so L=4. And now we can either say that F is the only letter unallocated, or look at the left-hand side of the first subtraction, or probably take various other routes, to get F=7.

The final result:

    0 1 2 3 4 5 6 7 8 9
D O U G L A S F I R

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

• We might have inferred $O=1$ from "everyone all over the globe (O)". Commented Jan 4 at 12:56

We try a(n almost) purely mathematical solution, without using the clues in the first part of Gareth's answer. We begin by noting that only nine different digits are specified in the long division, so we will have to infer the remaining digit from context. (If the message is the one indicated by Gareth, the D is not shown.)

From the second layer of subtractions in the tens and hundreds column of the dividend, we have

$$U\equiv 2S\bmod 10$$ (modulo 10 is understood from here on)

$$G\equiv 2S+1$$

G and U must be different; the +1 for the G comes from the fact that we had to borrow from the hundreds column to the tens column in this subtraction.

The third level subtraction in the tens column then gives either of the following, depending on whether we had to borrow from the tens column into the ones column (currently unknown):

$$S\equiv 2G$$

$$S\equiv 2G+1$$

If $$S\equiv 2G+1$$ and $$G\equiv 2S+1$$, then $$S\equiv 4S+3$$ from which $$S=9$$ and then $$G=9$$. This is contradictory because only one letter is allowed for each digit. Trying $$S\equiv 2G\not\equiv 2G+1$$ leads to $$S\equiv 4S+2$$, therefore $$S=6$$ and we infer $$G=3$$. We also have $$U\equiv 2S=2$$. Note that $$G, which figures into a later stage of the solution.

When these results are inserted, we may conclude from the second level subtraction that $$R=9$$ and from the third level subtraction that $$L=4$$.

So we know: $$U=2, G=3, L=4, S=6, R=9$$.

We now know seek letters $$A$$ and $$O$$ to complete the third level subtraction. These must add up to $$6$$ since $$G implies no borrowing into the tens column. From the known values of other letters we may infer that $$\{A,O\}=\{1,5\}$$, but in which order?

The subtrahend in the third level subtraction is known to be $$54132$$ or $$14532$$; either way it must be a single digit times a four-digit divisor. The single digit, of course, is greater than or equal to $$2$$ for $$14532$$ or greater than or equal to $$6$$ for $$54132$$.

Standard divisibility tests reveal that $$54132$$ is divisible only by $$6$$ among sufficiently large digits (there are smaller factors but these do not match with a four-digit divisor). Dividing $$54132$$ by $$6$$ yields $$9022$$, which does not admit a four-digit multiple ending in $$USS=266$$. So that possibility fails and we identify the third subtrahend as $$14532$$; thus $$O=1,A=5$$.

The number $$14532$$ is divisible by $$2,3,4,6,7$$ by standard testing. Division by $$7$$ gives $$2076$$, which again fails to give a four-digit multiple ending in $$266$$; ditto for $$14532/3=4844$$. But $$14532/2=7\color{blue}{266}$$, and division by $$4$$ or $$6$$ will also give $$7266$$ as a multiple of the quotient. Thereby $$F=7$$ and it becomes evident that $$I=8$$.

We have now rendered all the digits $$1$$ through $$9$$ appearing in the subtractions and thus have most of the hidden message:

_OUGLASFIR

The first letter, corresponding to $$0$$, is not given in the subtractions, but the only coherent choice (at least in English) is to use the letter D making the final answer

$$\color{blue}{\text{Douglas fir}}.$$

This solution was obtained without completely solving the division; the blanks in the dividend, divisor and quotient are not specified. Nonetheless, the division turns out to be unique. First note that with $$7266$$ as one of the subtrahends, the divisor must be a factor of that number and must also give the first subtrahend, $$58128$$, upon multiplication by a single digit. Only $$7266$$ (FUSS) itself satisfies these requirements. Then from the subtrahends the digits in the quotient must be $$812$$ (IOU). The remainder is already known to be $$2134$$. From these the dividend may be computed as $$5902126$$ (ARDUOUS).

• didn't expect to see this, but pretty cool! I wouldn't accept this answer without explanations of the clues (even setting aside the fact that there already is an accepted answer), but it's always neat to see different ways of solving the same problem. nicely done! Commented Jan 3 at 23:28
• It turns out you can recover the division elements too, see my addition. Commented Jan 3 at 23:44