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 2 of 2.

Murugan said:   1 decade ago
@Priyank is right but I think that raj method of identification is wrong because, how this is right== 7 (111) , 4 (100) , 8 (1000) total of 10 bits.

i.e. (1111001000) mark from LSB Then you feel the answer is wrong if you separate in hexadecimal then you have to separate by four bits then 0011 1100 1000==3c8. This is wrong answer. Any correction please reply.

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)

Papu said:   1 decade ago
2^9=512 that means we can write at most 512 with 9 digits.
2^19=1024 that means we can represent up to 1024 with 10 digits.

Bhavana said:   1 decade ago
Thanks pavan and raji. It was an easy way to explain.

Supraja said:   1 decade ago
Thank you raji. Thats quite easy method. Will it hold correct for similar question with different value.

Vinsan@hyderabad said:   1 decade ago
748= 512 + 0 + 128 + 64 + 32 + 0 + 8 + 4 + 0 +0(2 powers in order)
748= 10 1110 1100 Total ten bits required to represent 748 numbers

Naik said:   1 decade ago
8-4-2-1 method is easy to understand

Raji said:   1 decade ago
By using 8-4-2-1 method also we can find it..
like binary number of 7(111),4(100), 8(1000) total of 10 bits.. i.e(1111001000)

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


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