Applications of Bases: Part I, Problem 1

We note that the nth number is just n in binary! Therefore, 101101 is the 101101(base 2)th number. 101101(base 2)=45, so we conclude that it is the 45th number in the list.

$$ The series will form the first $n$ -term of binary sequence and the position of $101101$ is the sequence of $n$ -term in decimal number. Therefore, to determine the position of $101101_2$ all we have to do is just converting $101101_2$ to base $10$ . $$ $\color{#D61F06}{101101_2}=(\color{#D61F06}{1}\times2^5+\color{#D61F06}{0}\times2^4+\color{#D61F06}{1}\times2^3+\color{#D61F06}{1}\times2^2+\color{#D61F06}{0}\times2^1+\color{#D61F06}{1}\times2^0)_{10}=\boxed{\color{#3D99F6}{45}}$ $$

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$\Large\color{#3D99F6}{\text{\# }\mathbb{Q.E.D.}\text{ \#}}$

$\left ( 101101\right )_{2}=\left ( 1\times 2^{5} \right )+\left ( 0\times 2^{4} \right )+\left ( 1\times 2^{3} \right )+\left ( 1\times 2^{2} \right )+\left (0\times 2^{1} \right )+\left ( 1\times 2^{0} \right )$ $=32+0+8+4+0+1=45$

2^5 + 0 + 2^3 + 2^2 + 0 + 2^0 = 45 Uhm binary digits

101101(2)=45(10)

here nth no# is in binary no#...means base 2 so there decimal value identity it's place value or say position in chronological order of 0 and 1

Hello all,

as the binary series mentined as 1,10,11............,101101....., what is the the position 101101?

change those to decimals,

1,10,11......=1,2,3......,45....,i make them as an arithmetic series,

as for 1011011 = 45,

a=1, d=1, Tn=45,

Tn = a+(n-1)d

45 = 1+ (n-1)1

45 =1 + n -1

n=45,

Therefore, 101101 it is in 45th position in the series.....

1-1 10-2 11-3......................... 101101-45 here, sequence starts at 1 in 1st position and continued with a common difference of 1 up to 45. therefore, number position 1 1 10(2) 2 11(3) 3 .. .. 101101(45) 45

This question is as simple as tricky it looks if you have a sharp mind that can quickly get to work to the problem than you have no problem to solving it Using conversion of binary to decimal 101101=1x2^5+0x2^4+1x2^3+1x2^2+0x2^1+1x2^0=45 :P