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 1 of 2.
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
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)
like binary number of 7(111),4(100), 8(1000) total of 10 bits.. i.e(1111001000)
Naik said:
1 decade ago
8-4-2-1 method is easy to understand
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
748= 10 1110 1100 Total ten bits required to represent 748 numbers
Supraja said:
1 decade ago
Thank you raji. Thats quite easy method. Will it hold correct for similar question with different value.
Bhavana said:
1 decade ago
Thanks pavan and raji. It was an easy way to explain.
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.
2^19=1024 that means we can represent up to 1024 with 10 digits.
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.
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)
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.
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.
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.
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.
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