Question

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

Answer 1

Hint: The arithmetic mean (A.M.) of any given numbers is the sum of numbers divided by the total number of numbers so use this concept to reach the solution of the question.

Complete step-by-step answer:

The given set of 9 numbers is {9, 99, 999, ……………………….. 999999999}

The arithmetic mean of n numbers ${a_1},{a_2},{a_3}................{a_n}$is given as

${\text{A}}{\text{.M = }}\dfrac{{{a_1} + {a_2} + {a_3} + ...............{a_n}}}{n}$

It is given that the arithmetic mean of the nine numbers of the given set is N.

Thus the arithmetic of given set of 9 numbers will be

$N = \dfrac{{9 + 99 + 999 + 9999 + ............... + 999999999}}{9}$

Taking 9 common form the numerator part we get

${\text{N = }}\dfrac{{9\left( {1 + 11 + 111 + ........ + 111111111} \right)}}{9}$

$ \Rightarrow N{\text{ = 1 + 11 + 111 + 1111 + }}.................{\text{ + 111111111}}$

Now there are two ways to add this A.M first one is direct addition. So if we do direct

addition then

$
  {\text{N = 1 + 11 + 111 + 1111 + }}.................{\text{ + 111111111}} \\
   \Rightarrow {\text{123456789}} \\
$

Hence N = 123456789

In the second method we will depict the pattern in digits after addition of two consecutive numbers.

So let’s first add first two digits that is 1 + 11 = 12

Now add 12 to its successive digit in the series, thus 12 + 111 = 123

Now add 123 to its successive digits in the series, thus 123 + 1111 = 1234

Clearly we can see that in every addition 1 digit more is getting added in the final answer and that digit is the (+1) of the previous digit, so we can say that the next output will be 12345, and then it will continue.

Now since we have to add up only 9 times as there are only 9 digits in the set thus, the final output sum will be 123456789.

Thus from both the methods we can say that

N = 123456789

Hence it is clear that N does not contain 0 in it hence option (a) is right

Note: Whenever we face such types of questions the key point we have to remember is arithmetic mean whose statement is written above so according to this property we can get the required arithmetic mean of the given numbers as above we can use another method for addition of numbers as we see that 9, 99, 999 and so on is written as ((10 – 1), (100 – 1), (1000 – 1) and so on) so as we see that (10, 100, 1000 and so on) forms a geometric progression (G.P) whose first term (a) is 10, common ratio (r) is (100/10) =10, and number of terms(n) is 9 so the given series is written as [(10 +100 +1000 +…+100000000) – (1 +1 +1 +1 …….. + 1)] so we use the formula of sum (Sn) of G.P which is Sn = a[(rn - 1) / (r – 1)] where symbols have their usual meaning.