Question

let n be positive integer that \[\log_{2} \log_{2} \log_{2} \log_{2} \log_{2} n<0<\log_{2} \log_{2} \log_{2} \log_{2}n\] Let l be the number of digits in the binary expansion of n. Then the minimum and the maximum possible values of l are

Answer 1

@aroub ?
@bibby ?
@campbell_st ?

Answer 2

@Hero ?

Answer 3

@nincompoop ?

Answer 4

@sammixboo ?

Answer 5

\[\large
\log_{2} \log_{2} \log_{2} \log_{2} \log_{2} n<0<\log_{2} \log_{2} \log_{2} \log_{2}n \\
\large
\text{If } a \lt b \lt c, \text{ then }2^a < 2^b < 2^c. \normalsize ~~\text{Therefore, } \\
\large
2^{\log_{2} \log_{2} \log_{2} \log_{2} \log_{2} n} < 2^0 < 2^{\log_{2} \log_{2} \log_{2} \log_{2} n} \normalsize\text{ which simplifies to one less log:}\\
\large
\log_{2} \log_{2} \log_{2} \log_{2} n<1<\log_{2} \log_{2} \log_{2}n ~~~~
\normalsize \text{ since }~2^{\log_2(x)} = x\\
\text{Raise them again to the power of 2 and simplify. There will be one less log:} \\
\large
\log_{2} \log_{2} \log_{2} n<2^1=2< \log_{2} \log_{2}n \\
\text{Raise them again to the power of 2 and simplify. } \\ \large
\log_{2} \log_{2} n<2^2=4< \log_{2}n \\
\text{Raise them again to the power of 2 and simplify. } \\ \large
\log_{2} n<2^4=16< n. ~~~ \\
\text{n has to be > 16. This is the lower limit of 'n'} \\
\text{Raise them again to the power of 2 and simplify. } \\ \large
n < 2^{16} = 65536 < 2^n \\
\text{n has to be < 65536. This is the upper limit of 'n'} \\
\]Since n is an integer, the lowest value is 17 and the highest value is 65535.

Answer 6

17 in binary is: 1 0001 (5 binary digits)
65535 in binary is: 1111 1111 1111 1111 (16 binary digits)