Question

When ${{\left( 55 \right)}_{10}}$ is represented in base 25 then the expression is\[\]
A. ${{\left( 25 \right)}_{25}}$\[\]
B. ${{\left( 25 \right)}_{35}}$\[\]
C. ${{\left( 55 \right)}_{25}}$\[\]
D. None of these \[\]

Answer 1

Hint: We divide $N=55$ by the base $b$ and have quotient as ${{q}_{0}}$ and the remainder as ${{d}_{0}}$.We divide ${{q}_{0}}$ by the base $b$ and have quotient as ${{q}_{1}}$ and the remainder as ${{d}_{1}}$.We continue until we get ${{q}_{n}}=0$ for some natural number $n>1.$The equivalent number with base $b$ is ${{\left( N \right)}_{b}}={{d}_{n}}...{{d}_{2}}{{d}_{1}}{{d}_{0}}$.\[\]

Complete step-by-step solution
We know that form the expansion in number system of a number with $n$ digits with base $b$ is denoted as ${{\left( N \right)}_{b}}={{d}_{n}}{{d}_{n-1}}....{{d}_{1}}{{d}_{0}}$ where ${{d}_{n}},{{d}_{n-1}},....{{d}_{1}},{{d}_{0}}$ are the digits can be expanded as
\[\begin{align}
  & {{\left( N \right)}_{b}}={{d}_{n}}{{d}_{n-1}}....{{d}_{1}}{{d}_{0}} \\
 & ={{d}_{n}}\times {{b}^{n}}+{{d}_{n-1}}\times {{b}^{n-1}}+....{{d}_{1}}\times {{b}^{1}}+{{d}_{0}}\times {{b}^{0}} \\
\end{align}\]
The exponent terms of $b$ are called weights of digits. If the base is 10 it is called decimal system and we write ${{\left( N \right)}_{10}}$ as ,
\[{{\left( N \right)}_{10}}={{d}_{n}}\times {{10}^{n}}+{{d}_{n-1}}\times {{10}^{n-1}}+....{{d}_{1}}\times {{10}^{1}}+{{d}_{0}}\times {{10}^{0}}\]
If we want to convert a number in decimal numeral system say ${{\left( N \right)}_{10}}$to its equivalent number in the numeral system with base 10 . We use the following working rule,\[\]
1. We divide $N$ by the base $b$ and have quotient as ${{q}_{0}}$ and the remainder as ${{d}_{0}}$.\[\]
2. We divide ${{q}_{0}}$ by the base $b$ and have quotient as ${{q}_{1}}$ and the remainder as ${{d}_{1}}$.\[\]
3. We continue until we get ${{q}_{n}}=0$for some natural number $n>1.$\[\]
The equivalent number with base $b$ is
\[{{\left( N \right)}_{b}}={{d}_{n}}...{{d}_{2}}{{d}_{1}}{{d}_{0}}\]

We see that the give number is ${{\left( 55 \right)}_{10}}$. We have to find its equivalent in the numeral system with base 25. We follow the working rule.\[\]
1. We divide $N=55$ by the base $b=25$ and have quotient as ${{q}_{o}}=2$ and the remainder as ${{d}_{0}}=5$.\[\]
2. We divide ${{q}_{0}}=2$ by the base $b=25$ and have quotient as ${{q}_{1}}=0$ and the remainder as ${{d}_{1}}=2$.\[\]
The equivalent number with base 25 is
\[{{\left( N \right)}_{25}}={{d}_{1}}{{d}_{0}}=25\]
So the correct choice is A.

Note: We observe that the maximum number of symbols to represent a number is equal to $b$, for example in the decimal number system the maximum number of symbols we use is equal to 10 and the symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 which are also called digits. So if we find remainder $9 < {{d}_{i}} < 25$ where $i=0,1,...,n$ in the working rule then ${{d}_{i}}$ has to be represented by new unique symbol like English alphabet.