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!


While on vacation, respectable advisor can go tell King to invade valley in order to obtain fiddle

Oops! It looks like all three clues got - well, you read the title. Hopefully you can figure out how to unstick them.

Accessible version:


1 Answer 1


Answers to cryptic clues:

While on vacation, respectable advisor can...
R_E + A_R = REAR (can)

...go tell...
BID, double definition (a go is a bid/attempt, to tell someone is to bid them)

...King to invade valley in order to obtain fiddle
BB to invade DALE = DABBLE (fiddle)

Alphametic logic

B,D,R,T,U cannot be zero as they appear as starting digits. I cannot be zero because I×REAR=UCICB is not zero. From RCLEE-RITAT=RBU being a three-digit number we see that C=I+1, so C cannot be zero. From UEDAL-UCICB>0 we see E>C, so E cannot be zero. A cannot be zero because DABB-TTBA=UEDA doesn't end in B. This leaves only one option for zero, which is L.

Alphametic with L=0 filled

From UEDA0-UCICB=RC0E we see that 10-B=E, and that (A-1)-C=0 -> A=C+1. Since we know from earlier that C=I+1, that means that I,C,A must be three consecutive digits in ascending order.

DABB-TTBA=UEDA means that B is even (since either B=2A or 10+B=2A). From the last digits of UEDA0-UCICB=RC0E we also see that E is even. And from the last digits of RC0EE-RITAT=RBU we see that T,U are either both even or both odd. If they were both even we'd have too many even digits (L=0, B, E, T, U and at least one of I,C,A). So T,U are both odd.

The placement of I,C,A is restricted now. It can't be 1,2,3 or I×REAR would be REAR. It can't be 2,3,4 or the remaining even digits B,E would be 6,8 and there's no way to make 10-B=E. It can't be 3,4,5 or B would be 0, which is already taken by L. It can't be 5,6,7 or B would be 4, making E=6 which is already taken by C. It can't be 6,7,8 or B would be 6, which is taken by I. And it can't be 7,8,9 or there'd be no room left for E>C. So the only option left is I,C,A = 4,5,6.

Alphametic with L=0, I=4, C=5 and A=6 filled

6+6=10+B gives B=2, and since we know 10-B=E, we get E=8. All of a sudden it looks like we might know what we'll be spelling here...

Alphametic with additional B=2 and E=8 filled

From the last subtraction we see 8-T=U or 18-T=U. The only options that work are T,U being 1,7 in some order. So D,R are 3,9 in some order. R cannot be 9 since 2×R86R is a four-digit number. So R=3 and D=9.

Alphametic with additional R=3 and D=9 filled

To finish off we can just type in 2×3863 into a calculator, giving 7726 meaning T=7. This means U=1, giving the final answer. To get our clues unstuck, we need to get them

Solved alphametic


  • 5
    $\begingroup$ nicely done! it was very interesting to read your solve path considering it was completely different from mine when I was test solving $\endgroup$
    – juicifer
    Commented Aug 16, 2023 at 12:58

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