Question

Number of 9-digit numbers that are divisible by 9 and have all digits unique is $K\times 8!$. Find the value of K.

Answer 1

Hint: Use the fact that the sum of the first n natural number is given by $\dfrac{n\left( n+1 \right)}{2}$ which is divisible by n is odd. Hence prove that the nine digit number will be divisible by 9 if and only if the digits used are $0,1,2,3,4,5,6,7,8$ or $1,2,3,4,5,6,7,8,9$. Use the fact that the number of ways of arranging n different things in n places is $n!$. Hence find the number of 9-digit numbers with different digits divisible by 9. Hence find the value of K.

Complete step-by-step answer:
We know that the sum of the first n natural number is given by ${{S}_{n}}=\dfrac{n\left( n+1 \right)}{2}$. If n is odd, we have $n=2k+1$.
Hence, we have
${{S}_{n}}=\dfrac{\left( 2k+1 \right)\left( 2k+2 \right)}{2}=\left( k+1 \right)\left( 2k+1 \right)$ which is divisible by 2k+1 = n.
Hence the sum of the first n natural number is divisible by n if n is odd.
Hence, we have
$1+2+3+4+\cdots +9$ is divisible by 9.
We know that a number is divisible by 9 if the sum of its digits is divisible by 9. Hence the 9-digit numbers formed by using the digits $1,2,3,4,5,6,7,8,9$ taken all at a time will be divisible by 9.
Since 9 is divisible by 9, we have $1+2+3+\cdots +8$ is also divisible by 9.
Hence, we have
$0+1+2+\cdots +8$ is also divisible by 9.
Hence the 9-digit numbers formed by using the digits $0,1,2,3,4,5,6,7,8$ taken all at a time will be divisible by 9.
Since any of the other numbers 1-8 is not divisible by 9 no other such combination is possible.
Hence the nine digit number will be divisible by 9 if and only if the digits used are $0,1,2,3,4,5,6,7,8$ or $1,2,3,4,5,6,7,8,9$
Case I : Digits used are $0,1,2,3,4,5,6,7,8$
The leftmost place can be filled in 8 ways (0 cannot be placed there as it will make the number an 8-digit number).
The remaining 8 digits can be arranged in 8 places in 8! Ways.
Hence the total number of numbers that can be formed is $8\times 8!$
Case II: Digits used are $1,2,3,4,5,6,7,8,9$
The nine digits can be arranged in 9 places in 9! Ways.
Hence the total number of numbers that can be formed is $9!=9\times 8!$
Hence the total numbers which are divisible by 9 is $9\times 8!+8\times 8!=17\times 8!$
Hence, we have K = 17.

Note: [1] In solving the problem involving numbers that can be formed using certain digits, we must first fill those places which have the most restriction as it makes the calculation easy. In the above question, the highest value place was having restriction (0 cannot be placed there). That is why we filled the place first and then the rest.