How big of a dictionary would be required if my alphabet had 10 numbers?

Question

My language consists of 0/1/2/3/4/5/6/7/8/9 .
Zip codes identifies towns, villages and cities. if I read 43138. I would be saying: Logan, OH

Love it. The answers are great.

Punctuation is a part of all languages. It only takes a decimal point to start a chain letter. As in Pi (8)

Answer 1

The possibilities are infinite unless you impose limits on the size of your words.

For any word of N "letters" there would be 10^N possible combinations.

So:

There are 10 one letter words
There are 100 two letter words
There are 1000 three letter words
etc

Answer 2

Assuming you can only create 5 digit "words" (a.k.a. zip codes), you have 10 choices for each "letter".

10 x 10 x 10 x 10 x 10 = 10^5 = 100,000

The values would be 00000 through 99999 ---> 100,000 combinations.

Edit:
With your revised details where length is *not* restricted, the answer is an infinite number of combinations of digits. Your "dictionary" would just be the positive integers, of which there are an infinite number. Adding in a decimal place just gets you to the set of positive real numbers, which is also infinitely big. If you want to add a negative sign, you can include the negative numbers.

The answer is your dictionary would be *infinitely* big!

Answer 3

Are you including one-letter words? Two-letter words? ten-letter words? etc.?
There are 10 possible one-letter words : 1, 2, 3, 4, and so on.
There are 100 possible two-letter words: 00,01,02,03,04 . . .all the way to 99
There are 10 to the 3rd power (1000) possible three letter words, etc.
If you're asking about zip codes, then there are 100,000 possible zipcodes, all the way from 00000 to 99999.

If you're asking about all possible words from one-letter words to five-letter words, there are 111,110 words (that's 10+100+1000+10,000+100,000).

If you have a ten letter limit, there are 11,111,111,110 words (that's 10+100+1000+10,000+100,000+1,000,000 and so on to 10,000,000,000)