Computer Science - Digital Computer Electronics - Discussion
Discussion Forum : Digital Computer Electronics - Section 1 (Q.No. 40)
40.
How many bits are required to encode all twenty six letters ten symbols, and ten numerals?
Discussion:
4 comments Page 1 of 1.
Labeeba said:
4 years ago
6 bits are required to encode all 26 letters, 10 symbols, and numerals.
If we want to represent one character from the 26-letter Roman alphabet (A-Z), then we need log2(26) = 4.7 bits. We will need 5 bits.
If we want to represent one character from the 26-letter Roman alphabet (A-Z), then we need log2(26) = 4.7 bits. We will need 5 bits.
Moni said:
1 decade ago
26 letters + 10 symbols + 10 numerals = 46
any binary code is represented in powers of 2
2 power 6 =64 can represent the 46...
any binary code is represented in powers of 2
2 power 6 =64 can represent the 46...
(1)
S raj said:
8 months ago
32 16 8 4 2 1.
Where 32 ≤ 46.
Hence 6 ie 101110 is required.
Where 32 ≤ 46.
Hence 6 ie 101110 is required.
Ritu kumari said:
9 years ago
Thanks @Moni.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers