Question

The arithmetic mean of the nine numbers in the given set $\left\{ 9,99,999,999999999 \right\}$ is a 9 digit number N, all whose digits are distinct. The number N does not contain the digit.

Answer 1

Hint:We will be using the concepts of arithmetic to solve the problem. We will also be using the concept of geometric progression to find the sum of 9 numbers and simplify the solution.

Complete step-by-step answer:
Now, we have been given nine numbers as $\left\{ 9,99,999,999999999 \right\}$ we have to find the digit which is not in the mean of these 9 numbers.
Now, we know that the arithmetic mean of two numbers a and b is $\dfrac{a+b}{2}$. So, for 9 numbers will be,
$mean=\dfrac{9+99+999+......+999999999}{9}$
Now, we can write 9 as 10 – 1 and 99 as 100 – 1. Similarly, all the terms can be written in \[{{10}^{n}}-1\] form. So, we have,
$\begin{align}
  & mean=\dfrac{10-1+{{10}^{2}}-1+{{10}^{3}}-1+.....+{{10}^{9}}-1}{9} \\
 & =\dfrac{10+{{10}^{2}}+{{10}^{3}}+.....+{{10}^{9}}-9}{9} \\
\end{align}$
Now, we know that the sum of a geometric progression is,
$\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where r is common ratio.
n is the number of terms and a is the first term.
So, now we have,
$\begin{align}
  & mean=\dfrac{\dfrac{10\left( {{10}^{9}}-1 \right)}{10-1}-9}{9} \\
 & =\dfrac{\dfrac{10\left( 999999999 \right)}{9}-9}{9} \\
 & =\dfrac{10\left( 111111111 \right)-9}{9} \\
 & =\dfrac{1111111110-9}{9} \\
 & =\dfrac{1111111110}{9}-1 \\
 & N=123456790-1 \\
 & =123456789 \\
\end{align}$
Hence, the number does not contain zero digit. So, the correct answer is 0.

Note: To solve these types of questions it is important to note the way we have converted the sum of $9+99+999+......+999999999$ in a geometric progression and simplified it for the answer.