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Solution


Q.

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        [2023]
1 18   
2 21   
3 22   
4 20  

Ans.

(3)

Case I: All the three digits are same i.e., 111, 333, 555, 888. All the numbers are divisible by 3.

Case II: Two digits are same.

We can see that when two digits are same, then numbers having digits 5, 5, 8 or 8, 8, 5 will be divisible by 3 as their sum is divisible by 3.

Now, for each number, we have 3!2!=3 arrangements.

∴    In this case, we have 6 numbers divisible by 3.

Case III: All digits are distinct.

i.e., (1, 3, 5), (1, 3, 8)

So, number formed = 2 × 3! = 12

Hence, total 3-digit numbers formed by digits 1, 3, 5, 8 and divisible by 3 is 4 + 6 + 12 = 22.