Question

Find the radix scale in which numbers denoted by 479, 698, 907 are in Arithmetic Progression.

Answer 1

Hint: We will use the concept of number system and Arithmetic Progression to solve the above question. Let us suppose that number ${{\left( {{b}_{3}}{{b}_{2}}{{b}_{1}}{{b}_{0}} \right)}_{r}}$ is in radix scale r then to convert it into decimal number system we will get it as ${{b}_{3}}\times {{r}^{3}}+{{b}_{2}}\times {{r}^{2}}+{{b}_{1}}\times {{r}^{1}}+{{b}_{0}}\times {{r}^{0}}$. And, from arithmetic progression we know that when three numbers ‘a’, ‘b’, and ‘c’ are in AP then 2b = a + c.

Complete step by step answer:
We will use the concept of conversion of the radix system from the number system to solve the above question.
Let us suppose that we are given a number ${{b}_{3}}{{b}_{2}}{{b}_{1}}{{b}_{0}}$ where ${{b}_{0}},{{b}_{1}},{{b}_{2}},{{b}_{3}}$ are digits in radix system ‘r’.
Then, to convert the given number ${{\left( {{b}_{3}}{{b}_{2}}{{b}_{1}}{{b}_{0}} \right)}_{r}}$from radix scale ‘r’ to radix scale 10 we will use:
$\Rightarrow {{\left( {{b}_{3}}{{b}_{2}}{{b}_{1}}{{b}_{0}} \right)}_{r}}={{\left( {{b}_{3}}\times {{r}^{3}}+{{b}_{2}}\times {{r}^{2}}+{{b}_{1}}\times {{r}^{1}}+{{b}_{0}}\times {{r}^{0}} \right)}_{10}}$
And, from arithmetic progression we know that when three number ‘a’, ‘b’, and ‘c’ are in AP then relation between these number is given as:
2b = a + c
Since, we know from the question that 479, 698, 907 are in Arithmetic Progression and they are in some other radix scale not 10.
Let us suppose that 479, 698, 907 are in radix scale ‘r’ and they all are also in Arithmetic Progression.
Here, a = 479, b = 698, and c = 907
$\begin{align}
  & \Rightarrow 2b=a+c \\
 & \Rightarrow 2{{\left( 698 \right)}_{r}}={{\left( 479 \right)}_{r}}+{{\left( 907 \right)}_{r}} \\
\end{align}$
Now, since we can not solve it in the radix system ‘r’ simply, we will convert it into a decimal number system.
$\Rightarrow 2\left( 6\times {{r}^{2}}+9\times r+8 \right)=\left( 4\times {{r}^{2}}+7\times r+9 \right)+\left( 9\times {{r}^{2}}+0\times r+7 \right)$
$\Rightarrow 2\left( 6{{r}^{2}}+9r+8 \right)=\left( 4{{r}^{2}}+7r+9 \right)+\left( 9{{r}^{2}}+7 \right)$
Now, we will take 2 inside the bracket and simplify the RHS.
$\Rightarrow \left( 12{{r}^{2}}+18r+16 \right)=\left( 13{{r}^{2}}+7r+16 \right)$
Now, we will take all the term from RHS to LHS, then
$\Rightarrow \left( 12{{r}^{2}}+18r+16 \right)-\left( 13{{r}^{2}}+7r+16 \right)=0$
$\Rightarrow 12{{r}^{2}}+18r+16-13{{r}^{2}}-7r-16=0$
$\Rightarrow -{{r}^{2}}+11r=0$
Now, we will take minus common and then also ‘r’, then we will get:
\[\Rightarrow -r\left( r-11 \right)=0\]
\[\Rightarrow r\left( r-11 \right)=0\]
Now, we will equate each term to zero:
$\Rightarrow r=0,r-11=0$
$\Rightarrow r=0,11$
Since, we know that radix scale 0 is not possible so, r = 11 is the correct answer.
Hence, the radix scale in which numbers denoted by 479, 698, 907 are in Arithmetic Progression is 11.

Note:
Students are required to note that when we are given number in radix 10 and we have to convert it into any other radix ‘r’, then we follow the steps:
1). We will start dividing the given number by ‘r’ until we get remainder 0.
2). Then, we will store the remainder that we get in each iteration.
3). And, at last we will arrange the remainder in reverse order of which we have obtained.
Thus, the obtained number in the radix system ‘r’ will be equivalent to the number given in radix system 10.