PUZZLING Unlisted Numbers

The title means the powers-that-be won’t tell you what the number is; you have to find that out more deviously. Thus it’s an excellent choice to characterize this sort of alphametic crossword. So if your level of math anxiety is not too high, prepare to count on finger and toe, turn on brain and calculator, and go.



Instructions:

The clues are exercises in simple arithmetic. Every letter represents a number. A letter appearing more than once in a given clue will always stand for the same number; and two different letters in a clue can never mean the same number. By convention, no beginning letter can stand for a zero. When decoded into numbers, each clue makes a true arithmetical statement. Some clues have a unique solution, while others can be satisfied by several different arrangements of numbers. By using across entries to help with down and vice versa, the possibilities can be narrowed down to the final answer.

The clues are independent of one another so the same letter may take on different values in different clues – which is to say, the T in 7 across will very likely not have the same numerical value as the T in 8 across. (Note: O = letter O, 0 = number zero; X = letter X, x = multiplication times sign.)

A sample clue might be:

COO

+ COO

BABY;enter CABBY.

The object is to determine numerical values for each letter in CABBY and then enter the numbers in the diagram. Your reasoning might proceed thus:

(1) B= 1. If C had the largest value possible, 9, the biggest amount you could get for C + C, even with a carried 1, would still be under 20; so B must be less than 2. Being a beginning letter, B can’t be zero, so it must be 1.

(2) O = 5, Y = 0. Twice any number gives an even answer, so Y = 0, 2, 4, 6 or 8. But in the next column over, O + O comes out with a different ending number. The only way this could happen is with a carry, so O must be 5 or greater. Since the only difference between B and Y is a carry, B must just be one more than Y. Since you know B = 1, Y must be zero. And the only way Y can be zero is for O to be 5.

(3) Now you are left with: C + C plus a carry ends in A, and C itself is big enough that C + C creates a carry, so C must be 5 or greater. C can’t be 5 (O already is) so in table form:

C: 6 7 8 9

C+C+1→ A: 3 5 7 9

You can outlaw C = 7 and C = 9 because each leads to two different letters carrying the same value. So you have C and A equal to 6 and 3 or 8 and 7, respectively.

The outcome is that the letters in CABBY could stand for 63110 or 87110. So at this point, three numbers can be entered: 110; and the empty squares will await corroboration from the crossing clues.



Send the completed puzzle (or reasonable facsimile) to Puzzling, D Magazine, 1925 San Jacinto, Dallas, Texas 75201. All correct solutions will be held for one week after receipt of the first entry, at which time a drawing will take place to determine the winners. First winner will receive a $25 cash prize. Runner-up will receive a free one-year subscription to D. Winners and completed puzzle will appear in the September issue.