This is an interesting counting methods question that combines a very basic rule of test of divisibility of number by 3.

### Question

How many six digit positive integers comprising only the digits 1 and 2 can be formed such that the number is divisible by 3?

- 3
- 20
- 22
- 38
- 360

### Correct Answer

Choice C. 22 such numbers can be formed.

### Video Explanation

### Explanatory Answer

**Test of divisibility by 3**

If a number is divisible by 3, the sum of the digits of the number must be divisible by 3.

In this question we are forming 6 digit numbers comprising only two digits: 1 and 2.

We will list down various possibilities.

Case 1: 111111: All digits are 1s. The sum of the digits of the number is 6, which is divisible by 3. This number can rearrange in only **one way**.

Case 2: 222222: All digits of this number are 2s. The sum of the digits is 12, which is divisible by 3. This number also can rearrange in only **one way**.

Case 3: 111222: In case 1, where all digits are 1, the number was divisible by 3. If we remove 3 ones from 111111 and replace it with 3 twos, the resultant number will be divisible by 3.

This number can rearrange in = = 20 ways

Therefore, the total number of 6 digit numbers that can be formed using digits only 1s and 2s that are divisible by 3 = 1 + 1 + 20 = 22.

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