Digital Electronics - Digital Concepts - Discussion
Discussion Forum : Digital Concepts - General Questions (Q.No. 41)
41.
How many binary bits are necessary to represent 748 different numbers?
Discussion:
19 comments Page 2 of 2.
Arun said:
1 decade ago
@Priyank is very correct. Its standard method to find out binary to decimal. It will take less time too.
Pavan said:
1 decade ago
2^9=512, 2^7=128, 2^10=1024, 2^8=256
Therefore 10 bits are necessary to represent 748 different numbers
Therefore 10 bits are necessary to represent 748 different numbers
Markin said:
2 years ago
@All.
You can also take a log to the base 2 of 748 and you get approximately 10 bits.
You can also take a log to the base 2 of 748 and you get approximately 10 bits.
(1)
Rohit said:
10 years ago
Best and easy way is convert number into binary and count no. of bits required.
Faheem said:
10 years ago
Simple is that 748 is the decimal number so binary bit will be 10.
Rohit kk said:
7 years ago
748/16=(2EC)16,
(1011101100)2,
The number of digit =10.
(1011101100)2,
The number of digit =10.
Bhavana said:
1 decade ago
Thanks pavan and raji. It was an easy way to explain.
Avinash said:
5 years ago
8-4-2-1 process is used for binary to BCD conversion.
Naik said:
1 decade ago
8-4-2-1 method is easy to understand
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