Question

What is the decimal equivalent of the 25 digits of hexadecimal number \[{{\left( 100....001
\right)}_{16}}\]?
(a) \[{{2}^{23}}+1\]
(b) \[{{2}^{24}}+1\]
(c) \[{{2}^{92}}+1\]
(d) \[{{2}^{96}}+1\]

Answer 1

Hint: Assume 100…..001 as a number with the rightmost 1 as its unit place digit. Start from 0 and assign
numbers 1, 2, 3, …… till you reach the leftmost 1. Now, write \[{{\left( 100....001 \right)}_{16}}\] as: -
\[1\times {{16}^{0}}+0\times {{16}^{1}}+0\times {{16}^{2}}+.......+0\times {{16}^{22}}+0\times
{{16}^{23}}+1\times {{16}^{24}}\]. Now, simplify the above expression by converting 16 into \[{{2}^{4}}\],
to get the answer.

Complete step by step answer:
We have to write the decimal equivalent of the given hexadecimal number, \[{{\left( 100....001
\right)}_{16}}\].
We may note that there are 25 digits in 100…..001 in which there are twenty – three 0’s and two 1’s.
There 1’s are present at the left most and right most positions while 0’s are in between these 1’s.
Suppose we have a hexadecimal number of the form, \[{{\left( abcd \right)}_{16}}\], where ‘a’, ‘b’, ‘c’
and ‘d’ are four digits. We have to convert this hexadecimal number into decimal equivalent. What we
do is, we assign 0 to d, 1 to c, 2 to b and 3 to a. So, to convert into decimal equivalent, we have the
expression: -
\[\Rightarrow {{\left( abcd \right)}_{16}}=d\times {{16}^{0}}+c\times {{16}^{1}}+b\times
{{16}^{2}}+a\times {{16}^{3}}\]
Here, 16 is converted into \[{{2}^{4}}\]. So,
\[\Rightarrow {{\left( abcd \right)}_{16}}=d\times {{2}^{0}}+c\times {{2}^{4\times 1}}+b\times
{{2}^{4\times 2}}+a\times {{2}^{4\times 3}}\]
Now, similarly when we will assign 0, 1, 2, …… to (100….001) and write \[{{\left( 100....001
\right)}_{16}}\] into its decimal equivalent, we get,
\[\Rightarrow {{\left( 100...0001 \right)}_{16}}=1\times {{16}^{0}}+0\times {{16}^{1}}+0\times
{{16}^{2}}+.......+0\times {{16}^{22}}+0\times {{16}^{23}}+1\times {{16}^{24}}\]
Converting 16 into \[{{2}^{4}}\], we have,
\[\begin{align}

& \Rightarrow {{\left( 100...0001 \right)}_{16}}=1\times {{2}^{0}}+0\times {{2}^{4\times 1}}+0\times
{{2}^{4\times 2}}+.......+0\times {{2}^{4\times 22}}+0\times {{2}^{4\times 23}}+1\times {{2}^{4\times 24}}
\\
& \Rightarrow {{\left( 100...0001 \right)}_{16}}=1+{{2}^{96}} \\
& \Rightarrow {{\left( 100...0001 \right)}_{16}}={{2}^{96}}+1 \\
\end{align}\]

So, the correct answer is “Option D”.

Note: One may note that we have to start from 0 and always from the rightmost digit. The power is
increased one – by – one. Never start from the left most digits as it will give the wrong answer.
However, even after applying the wrong process here you will get the correct answer because
(100….001) is symmetric. But remember that every time this reverse process will not be applicable.