This is highly insecure.
For instance, if you see the word
guyk in the ciphertext, what could the corresponding plaintext word be?
With your scheme (where each letter is enciphered by adding a number between 0..9 to it modulo 26), there are only 139 English words that could have led to it. (Those 139 possibilities are things like
york: all of the words where the first letter is in the range
a-g, where the 2nd letter is in
j-u, the third letter is in
p-y, and the fourth letter is in
If you had used a one-time pad (where each number is enciphered by adding a number between 1..26 to it modulo 26), all four-letter English words would be possible. In particular, there would be 13208 possibilities for the corresponding plaintext word.
Thus your scheme leaks information that greatly narrows down the number of words, from 13208 possibilities to just 139 possibilities.
unbreakable + 59460927644 =
zwfxejmhhpi. If an attacker sees the ciphertext
zwfxejmhhpi, he knows that the plaintext word must have the form
q-z n-w w-f o-x v-e a-j d-m y-h y-h g-p z-i. There are only four English words that match that pattern: namely,
uncoachable. This gives the attacker a lot of information about the plaintext. The attacker may be able to figure out which word is most likely based upon context. Or, based upon the other candidates for other possible words in the sentence, the attacker may be able to recognize which word is correct.
If an attacker had the encipherment of an entire sentence, I suspect there would be only one plausible decryption, and it would be easy to find it by piecing together the possible decryptions of each word until you find something that makes sense as a sentence. If an attacker had the encipherment of an entire document, I suspect it would be easy for the attacker to recover much or all of the document.
In short, the scheme is highly insecure.