Digital Electronics - Number Systems and Codes - Discussion
Discussion Forum : Number Systems and Codes - General Questions (Q.No. 19)
19.
How many binary digits are required to count to 10010?
Discussion:
8 comments Page 1 of 1.
Deepika said:
1 decade ago
Can you give me the explanation about this question.
Ashit said:
1 decade ago
100 divide by two 50 reminder o
50/2 25 0
25/2 12 1
12/2 6 0
6/2 3 0
3/2 1 1
1
total seven bit 1100100
50/2 25 0
25/2 12 1
12/2 6 0
6/2 3 0
3/2 1 1
1
total seven bit 1100100
(5)
Jinesh Sir said:
1 decade ago
8 4 2 1 numbering system can applied upto max. 15. i.e. 4 Binary Input.
16 8 4 2 1 numbering system can applied upto max. 31. i.e.5 Binary Input.
32 16 8 4 2 1 numbering system can applied upto max. 63. I. E. 6 Binary Input.
64 32 16 8 4 2 1 numbering system can applied upto max. 127. I. E. 7 Binary Input.
Here 100 number is given and it is covered by 7 Binary digit as Input. Hence Answer is A.
16 8 4 2 1 numbering system can applied upto max. 31. i.e.5 Binary Input.
32 16 8 4 2 1 numbering system can applied upto max. 63. I. E. 6 Binary Input.
64 32 16 8 4 2 1 numbering system can applied upto max. 127. I. E. 7 Binary Input.
Here 100 number is given and it is covered by 7 Binary digit as Input. Hence Answer is A.
(6)
Ganesh kolipaka said:
8 years ago
Nice explanation, Thanks @Jinesh, @Ashit.
Jaga said:
8 years ago
Good explanation @Jinesh sir.
Sai said:
8 years ago
To count 100, 2^6 is the max available so from 2^0 to 2^6. Totally 7 bits.
Jeni angel said:
6 years ago
Please explain this in detail.
Sejal said:
5 years ago
Thank you all for explaining the answer.
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