Question

In How many six digit numbers are possible by using the digits
1.2.3.4.5.6.7 without repetition such that they are divisible
by 12

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Answer 1

Since, there is no repetition, numbers 1 to 6 will always be present in the 6 digit number.

We will go through this step by step.

Divisible by 1 and 1 at the unit place:

_ _ _ _ _ 1

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 2 and 2 at the unit place:

_ _ _ _ _ 2

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 3 and 3 at the unit place:

_ _ _ _ _ 3

Since, any number will have all the digits from 1 to 6 and the sum 1 + 2 + 3 + 4 + 5 + 6 = 21 is divisible by 3

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 4 and 4 at the unit place:

_ _ _ _ _ 4

Here, there are two cases

_ _ _ _ 2 4

This gives 4 x 3 x 2 x 1 = 24 numbers.

_ _ _ _ 6 4

This gives 4 x 3 x 2 x 1 = 24 numbers.

This gives us total of 24 + 24 = 48 numbers.

Divisible by 5 and 5 at the unit place:

_ _ _ _ _ 5

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

Divisible by 6 and 6 at the unit place:

_ _ _ _ _ 6

As all 6 digit numbers formed with 1 to 6 digits(without repetition) are divisible by 3 and numbers with 6 at the unit place are even.

This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers.

None of these cases will have numbers overlapping with each other.

So, Total numbers = 120 + 120 + 120 + 48 + 120 + 120 = 648

I know, there might be a cleverer way to solve the same problem, but this is the simplest way to solve it.

The only way this would not be the case, assuming we are working in base ten, is if the tens digit is odd and the ones digit is 4. That gives us 6!-3*4!=648 ways.