# Cryptogram

Cryptogram used to refer to a ciphertext . Nowadays a cryptogram or alphametics describes a mathematical puzzle . It is a mathematical equation or a system of equations of unknown numbers whose digits have been replaced by letters. The goal is to find the value of each letter. The equations are usually based on simple arithmetic operations such as addition , subtraction , multiplication or division .

A symbol puzzle that uses non-alphabetic symbols instead of letters is commonly known as a cryptogram.

## example

The classic example in English is:

```    S E N D         9 5 6 7
+   M O R E     +   1 0 8 5
-----------     -----------
= M O N E Y     = 1 0 6 5 2
```

The solution to this riddle is:

S = 9, E = 5, N = 6, D = 7, M = 1, O = 0, R = 8 and Y = 2.

## Solution of cryptograms

Solving cryptograms by hand is a mixture between testing and ruling out possibilities. For example, the following considerations and exclusions solve the SEND + MORE = MONEY cryptogram listed above (numbering the columns from right to left).

1. M = 1 , because it is the only way to carry over the sum of two digits from column 4 to column 5.
2. To get a carry over from column 4 to column 5, S = 8 or 9 , S + M = 9 or 10 and O = 0 or 1 . But since M = 1 , O = 0 must be.
3. If there were a carryover from column 3 to column 4, then E = 9 and N = 0 . But since O = 0 , there is no carry and S = 9 .
4. If there was no carryover from column 2 to column 3, then E = N , which is impossible. So there is a carry and N = E + 1 .
5. If there was no carryover from column 1 to column 2, then N + R = E mod 10 . With N = E + 1 it follows that E + 1 + R = E mod 10 and thus R = 9 . But since S = 9 , there is a carry and R = 8 .
6. To get a carry over from column 1 to column 2, D + E = 10 + Y must be. Since Y ≠ 0 or 1 , D + E ≥ 12 . If D = 7 , then E must be ≥ 5 . If N ≤ 7 and E = N - 1 , then E ≤ 6 . Hence E = 5 or 6 .
7. If E = 6 , then D = 7 . But since N is also 7 because of N = E + 1 , E = 5 . So N = 6 .
8. Since D + E ≥ 12 , D = 7 and therefore Y = 2 .

But you can also solve cryptograms in a purely logical way, without testing and trying.

## Examples of simple German cryptograms

Cryptogram Solutions) example
ONE + ONE = TWO 12 solutions 1407 + 1407 = 2814
TWO + TWO = FOUR 12 solutions 1397 + 1397 = 2794
ONE + FOUR = FIVE 24 solutions 9406 + 3495 = 12901
TWO + FOUR = SIX 12 solutions 8624 + 3427 = 12051
FOUR + FOUR = EIGHT 77 solutions 1345 + 1345 = 2690
ONE + EIGHT = NINE 168 solutions 2948 + 1306 = 4254
ONE + NINE = TEN 6 solutions 2930 + 3283 = 6213