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If you have 5 bits: how many distinct binary numbers can you represent. and what is there range?

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Submitted : 2018-06-15 15:16:51    Popularity:     

Tags: binary  distinct  bits  range  represent  

please tell me how you solved this too it would mean a lot

Answers:

2^5 is the distinctiveness, that's 32 of total, range would be about the signage, unsigned would be 0 to 31; while signed goes -16 to 15.

With n bits, there are 2^n different bit patterns. For five bits, that means 2^5=32 different bit patterns.

You can't state a number range, though, without first stating a "number format". The question probably wants "0 to 31" as an answer, interpreting each pattern as a five-place number in base 2.

Well-known signed formats would have different ranges. Two's complement, for example, would have a range of -16 to +15. Either one's complement or sign-and-magnitude would have a slightly smaller range of -15 to +15 (since each has separate representations for +0 and -0.)

5 bits means from 00000 to 11111. Each placeholder carries a weight of 2 raised to the placeholder's position from right to left beginning with 0 for the first placeholder. So the first placeholder has a weight of 2 raised to the 0 power which equals 1. The second placeholder (number 1) is 2 raised to the power of 1 which equal 2. So we have from left to right as weights 16, 8, 4, 2, 1. So adding these up 16 + 8 + 4 + 2 + 1, we get 31, but counting the zero it's actually a range of 32 (0 - 31).

The range is frolm 00000 to 11111 (binary).

2 power of 5 = 32. Same calculation say as if you had 3 denary numbers, that would be 10 power 3 = 1000.
The range depends. Could be 0 to 31 (unsigned number) or -16 to +15 (2's complement signed number). Or you could use each code to represent any number you like e.g. 00000 = 65432, 00001 = 7999999, 00010 = 345 etc. So the range is meaningless in that case.



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