Question

Suppose there is a special language containing 624 characters and it is decided to encode all characters of this language by using baseeight (i.e., octal) numbers with possible digits of 0, 1, 2, 3, 4, 5, 6, and 7 in each digit position. Then, the minimal number of octal digits to encode this language is

_____________________
. If it starts with all '0' in each digit position to encode the
first character and continues on by using every consecutive octal numbers until the last character of the language is encoded, the the last character
should be encoded as __________

Answer 1

The minimal number of octal digits required to encode a language with 624 characters is 4. The last character would be encoded as '1167' in octal, starting from '0000' and incrementing by one for each new character.

Step-by-step explanation:

To find the minimal number of octal digits needed to encode a special language containing 624 characters, we need to determine the smallest power of 8 that is greater than or equal to 624. This is because each octal digit can represent 8 different values (0-7), and with each additional digit, the number of possible combinations is multiplied by 8.

The powers of 8 are 1 (8^0), 8 (8^1), 64 (8^2), 512 (8^3), etc. So, we can see that 8^3 (512) is too small, but 8^4 (4096) is more than enough. Therefore, the minimal number of octal digits needed is 4.

To find the encoding for the last character, we start by encoding the first character with all zeroes '0000' in octal. Since there are 624 characters, the last character would be the 623rd increment (as we start counting from 0). In octal, this means adding 623 (decimal) to the initial '0000', which gives us '1167' in octal as the encoding for the last character.