Michael Linton's Bayeux Tapestry: 1066 - A Medieval Mosaic and Puzzles

Tri-Alphametic Puzzles

Each Alphametic puzzle is constructed from up to 10 letters. Each letter represents a number between 0 and 9. Two letters can not represent the same letter. The objective is to determine the number used to exchange for each letter. The only obvious clue is that a word can not start with a zero.

A good first step is to look for the number 0. Since the conventions outline that no word may start with 0, we know that neither A, C or N are 0.

Next we attempt to establish whether any columns add up to more than 10 (e.g. 4 + 7 = 11), which indicates this column carries over to the next column. From the example, column 1 contains only the letter N. For this to happen, the addition from the preceding column must have generated a carry, therefore N = 1. From column 4, if N + N = D, this means that D = 2.

From column 2, C = 9 as any number less than 9 would not have produced a carry to column 1. By the same reasoning, column 3 must also produce a carry. Using this we can determine that E = 0 from column 2.

Finally, in column 3, A + A = 0. This means that A = 5 as we have already determined that column 4 had no carry. We have now replaced all the letters with numbers, providing the solution: A=5, C=9, D=2, E=0, N=1.

Issue 2 Cover