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?
9
7
10
8
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 1 of 2.

Priyank said:   1 decade ago
748/2= div-374 rem=0
374/2= div-187 rem=0
187/2= div-93 rem=1
93/2= div-46 rem=1
46/2 div-23 rem=0
23/2 div-11 rem=1
11/2 div-5 rem=1
5/2 div-2 rem=1
2/2 div-1 rem=0
1/2 div=0 rem=1
Hence (748)=1011101100
So 10 digits are required to represent 748.
(5)

Panchakshari S H M said:   2 years ago
For 8 all 1's means it is 255,
For 7 all 1's means it is 127,
For 9 all 1's mean it is 511,
For 10 all 1's means, it is 1023.

So, the answer is 10.
(1)

Markin said:   2 years ago
@All.

You can also take a log to the base 2 of 748 and you get approximately 10 bits.
(1)

Sirisha said:   1 decade ago
Agreeing with @Murugan.

Example: Take 99.

If 9(1001),9(1001) total of 8 bits.

But 99=1100011, see 7 bits are enough.

@Raji method of identification is wrong.

Avinash said:   5 years ago
8-4-2-1 process is used for binary to BCD conversion.

Rohit kk said:   7 years ago
748/16=(2EC)16,
(1011101100)2,
The number of digit =10.

Nisha said:   9 years ago
@Priyank. You are correct because it is very easy to understand these binary to decimal conversion method.

Faheem said:   10 years ago
Simple is that 748 is the decimal number so binary bit will be 10.

Rohit said:   10 years ago
Best and easy way is convert number into binary and count no. of bits required.

Arun said:   1 decade ago
@Priyank is very correct. Its standard method to find out binary to decimal. It will take less time too.


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