Question

The number of four digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 which are divisible by 3, when repetition of digits is allowed any number of times.
A. 2187
B. 1458
C. 6561
D. 2916

Answer 1

Hint: To solve this problem, we should know the basic knowledge of combinations and Euclid division lemma. We know that any positive integer can be represented in the form of $3k,3k+1,3k+2$. This result comes from the Euclid division lemma which states that any positive integer can be written in the form $a=bq+r$ where $0\le r
Complete step by step answer:
To solve this question, let us assume that the four digit number is $abcd=1000a+100b+10c+d$.
We know that the divisibility rule for 3 is that a number $abcd$ is divisible by 3 when $a+b+c+d$ is divisible by 3.
To get to the answer, we can select the values of a, b, c independently and the values of d based on the value $a+b+c$.
We know the Euclid division lemma which states that any positive integer can be written in the form $a=bq+r$ where $0\le rWe know that any positive integer can be represented in the form of $3k,3k+1,3k+2$.
Let us assume that the value of $a+b+c=3k$. If we add $d=3\text{ or 6 or 9}$, we get
$\begin{align}
  & a+b+c+d=3k+\left( 3\text{ or 6 or 9} \right) \\
 & a+b+c+d=3{{k}_{1}} \\
\end{align}$
We can infer that the value of d can be $d=3\text{ or 6 or 9}$ to get the sum $a+b+c+d$ which is divisible by 3.

Let us assume that the value of $a+b+c=3k+1$. If we add $d=2\text{ or 5 or 8}$, we get
$\begin{align}
  & a+b+c+d=3k+\left( 2\text{ or 5 or 8} \right) \\
 & a+b+c+d=3{{k}_{2}} \\
\end{align}$
We can infer that the value of d can be $d=2\text{ or 5 or 8}$ to get the sum $a+b+c+d$ which is divisible by 3.
Let us assume that the value of $a+b+c=3k+2$. If we add $d=1\text{ or 4 or 7}$, we get
$\begin{align}
  & a+b+c+d=3k+\left( 1\text{ or 4 or 7} \right) \\
 & a+b+c+d=3{{k}_{3}} \\
\end{align}$
We can infer that the value of d can be $d=1\text{ or 4 or 7}$ to get the sum $a+b+c+d$ which is divisible by 3.
So, we can finally write that for every selection of the three values of a, b, c, we get 3 possibilities of the value d. The individual selections of a, b, c are independent to each other and there are 9 possibilities for each of them as repetition is allowed. So, we can write the total number of ways as
$\begin{align}
  & \text{Total ways}=\left( \text{No of a} \right)\times \left( \text{No of b} \right)\times \left( \text{No of c} \right)\times \left( \text{No of d} \right) \\
 & \text{Total ways}=9\times 9\times 9\times 3=729\times 2187 \\
\end{align}$
$\therefore $The total number of ways is 2187. The answer is option-A.

Note:
 Some students try to write case by case for the possible values of sum and write cases for each of them. For example, when $sum=6$, the possible cases are $\left\{ 1122,1212,2112,2211,1221,2121..... \right\}$. This will also get to the answer but the process is very lengthy and we can see that the answer is a large number to count. So, we should use the basic trick that for every selection of a, b, c, we get 3 possibilities of d.