Cryptorithms are puzzles where a single digit number replaces every letter, and the challenge is to find the corresponding numbers. For example:
These puzzles test your knowledge of addition properties (or other arithmetic operations), often with a not so hidden messageÉ
One way to solve this problem is to begin with the left most digit in the answer. In this case, L = 1 because the most that could be grouped (when the digits in a column sum to 10 or more) from the ten thousands column (the B column) is 1. At this step, you can also conclude that B = 9 and O = 0.
If E + C were the next two larges available digits, 8 + 7 = 16, and if we add 2 (the largest number that could be grouped from a previous column) then the sum is 18 < 20, which means B = 9 to produce L = 1. Furthermore, O = 0.
Now lets fill in our progress (notice I put in the group from the thousands column):
B = 1; L = 1; O = 0
At this point, the problem gets interesting because the solution is not unique (yes, sometimes there is more than one possible answer!). There are a lot of SÕs in the problem, so that might be a good place to go next. Lets look at some different possibilities for the oneÕs column and keep in mind, S + S + X = S, so S + X = 10.
Try I: S = 7; X = 3
If this were the case, the E + C must be 16 in order to add a group of 10 to the ten thousands place (the 9 and 1). This canÕt happen though because the largest two available numbers are 8 and 6, 8 + 6 = 14. Even if one were added to this, we still couldnÕt get 16É
Try II: S = 6; X =4
Here, E + C > 10 because earlier we showed that a group of 10 went with the 9. In fact E + C = 15. If E = 8, the A + U = 10, but this canÕt happen because the available digits would be 5, 3, and 2, no two of which sum to ten. So what if E = 7? Again this wonÕt work for similar reasons.
Try III: S = 2; X = 8
In this case, E + C = 11. LetÕs try E = 4 and C = 7. This implies that A + U = 11. One possibility is A = 5 and U = 6, leaving R = 4.
B = 9 C = 7 S = 2 L = 1
E = 4 U = 6 O = 0 O = 0
A = 5 B = 9 X = 8 S = 2
R = 4 S = 2 S = 2 E = 4
S = 2 R = 4
S = 2
Here is one answer, can you find any more???
If you would like to try some cryptorithms, press here.
Finding more answers. On the linked page, there are crytorithms using different operations, and these could be explored. These could be used in a classroom as an introduction to logic that may interest students. There are many conditional situations that result in these problems and force students to consider a number of different cases.