PUZZLING

By popular request from the mathematical minds in our city, here is another arithmetical square. The last time I slipped one of these in, the response indicated that a whole new audience had been tapped. The word wizards threw up their hands while the numbers nuts seized their chance. And the deluge of correct solutions proved that it’s really not all that hard. So go ahead, wordies, and give it a try this time. You’ve nothing to lose but your sanity and your self-respect.

Instructions:



Each clue represents an exercise in simple arithmetic. Every letter represents a number. The same latter appearing more than once in a clue always represents the same number. Conversely, two different letters in a clue cannot represent the same number. (And by convention, the beginning letter never represents a zero.) Each clue, once decoded into numbers, represents a true arithmetical statement. (Note: O = letter O,0 = numerical zero.)



The clues are entirely independent of each other, so the same letter can take on different values in different clues. That is, the T in 7 across may not have the same numerical value as the T in 8 across.

A sample clue:

DONALD + GERALD – ROBERT; enter ABORT

If it will help you to columnize each equation,do so:DONALD

+ GERALD

ROBERT



The object is to determine the numerical values for each of the letters in ABORT and enter that value in the diagram. For the sake of an easy starting point, suppose you have determined from the solution of another clue that the space in the diagram where the T in ABORT will be entered contains a zero, and therefore T=O. How do you proceed?

(1) D + D = 0. D cannot equal zero because already T=0, so D – 5. Thus D + D=-10; mark the0 in the column for the numerical value of T andcarry the 1.

(2) In the next column, L + L + l=R or R + 10(the 1 is the carried 1, the R + 10 indicates thepossibility that L + L + l equals a two-digit figure). Also R will be odd since L + L must beeven.

(3) In the fifth column, with a pattern likeO + E=O, there are two possibilities for E: zeroif there were no prior carried 1, or 9 if there werea carry. But since zero has already been shownto be the value of T, E=9. Furthermore, N + Rwith or without a carried 1 must be greater than9. Also, D + G + 1 = R, or G + 6 = R.

(4)Since we know that R is odd, it could be7 or 9 (it must be greater than 6). But E-9already, so R = 7. And therefore G = 1.

(5) Substituting in (2), we haveL+L + 1=R orR + 10=7orl7. Therefore, L + L-6 or 16 and Lmust equal 3 or 8. Since A + A = E=9, there mustbe a carried 1 from the prior column, so L + L =16(since it must be greater than 10) and L = 8.

(6) A + A + l=9, so A + A=8 or 18 and A=4 or9, Since E=9 already, A=4 and there is nocarry to the next column to the left.

(7) We are now left with the letters O, N, and Bunassigned, and the numbers 2, 3, and 6 stillavailable. We can make a table of the possibilities: N=2, 3, or 6. N + R=N + 7=BorB + 10.B=9, 0, or 3. We eliminate the first two possibilities for B because E=9 and T=0 already.Therefore B= 3, N = 6, and O = 2 and all lettershave been given their numerical equivalent.Applying these equivalents to the letters of theword ABORT, the final solution to be entered inthe diagram is 43270.

A caution to the solver: some clues can be satisfied by several different arrangements of numbers while some have a unique solution. By using across entries to help with down clues and vice-versa, the possibilities can be narrowed down to the final answer.

Note that 1 down and 5 down are multiplications and that 2 down is a subtraction.

Send the completed puzzle with name and address to Puzzling, D Magazine, 2902 Carlisle, Dallas 75204. 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 $26 cash prize. Runner-up will receive a free one-year subscription to D. Winners and completed puzzle will appear in the January issue.