The total number of three-digit numbers, divisible by 3, which can be formed using the dig…

Question

The total number of three-digit numbers, divisible by 3, which can be formed using the digits $1,3,5,8$, if repetition of digits is allowed, is :

Accepted Answer

The number of three-digit numbers divisible by 3 by considering the possible sums of digits that are divisible by 3. Your approach is as follows:

1. Sum of digits is 3: $(1, 1, 1)$ - 1 possible number

2. Sum of digits is 9: $(1, 3, 5)$ and $(3, 3, 3)$ -

Let's consider the cases separately :

a. Sum of digits is 9: $(1, 3, 5)$
   For this case, we can arrange the digits in $3!$ ways :
   135, 153, 315, 351, 513, and 531.

b. Sum of digits is 9: $(3, 3, 3)$
   For this case, since all the digits are the same, there is only 1 possible number :
   333.

Now, the total number of possible numbers when the sum of digits is 9 is :

$3! + 1 = 6 + 1 = 7$

3. Sum of digits is 12: $(1, 3, 8)$ - $3!$ possible numbers

4. Sum of digits is 15: $(5, 5, 5)$ - 1 possible number

5. Sum of digits is 18: $(5, 5, 8)$ - $\frac{3!}{2!}$ possible numbers (since 5 is repeated)

6. Sum of digits is 21: $(5, 8, 8)$ - $\frac{3!}{2!}$ possible numbers (since 8 is repeated)

7. Sum of digits is 24: $(8, 8, 8)$ - 1 possible number

Adding up the possible numbers for each case, we get:

$$
1 + 7 + 3! + 1 + \frac{3!}{2!} + \frac{3!}{2!} + 1 = 1 + 7 + 6 + 1 + 3 + 3 + 1 = 22
$$

So, there are a total of 22 three-digit numbers divisible by 3 that can be formed using the digits $1, 3, 5, 8$ with repetition allowed.

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