why is the unsigned binary range of 2^6 63 bits? -


unsigned 6 bits ranges 0 63, , can there because 2^6 = 64. wouldn't 6 bit binary string go 2^5? first bit 2^0 no ...? going wrong here? i'm aware simple question please me understand not know much

a 6 bit binary string goes 0b000000 to

  0b111111 = 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 =  32 +  16 +   8 +   4 +   2 +   1 =  63 =  64 - 1 = 2^6 - 1

that sum of 2^i i in 0, 1, … (n-1) evaluates 2^n - 1 peculiarity of number two.


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