Formations of Special 6-Digit Numbers
How many six digit numbers can be formed using the digits 1 to 6,
without repetition such that the number is divisible by the digit at its
unit place?
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Counting The Formations of 6-Digit Special Numbers
You may need to read the question first!
Let's remind that the numbers can't be repeated. So mathematically there are total 6! (720) unique numbers can be formed.
The number XXXXX1 will be always divisible by 1; so there we have 5! = 120 numbers.
The number XXXXX2 will be always divisible by 2; so there we have 5! = 120 numbers.
Since sum of all digits is 21 which is divisible by 3; the number XXXXX3 will be always divisible by 3. So we have 5! = 120 more such numbers.
The number XXXXY4 is divisible only when Y = 2 or 6. So in the case we have 2 x 4! = 48 numbers.
The number XXXXX5 will be always divisible by 5; so there we have 5! = 120 numbers.
The number XXXXX6 will be always divisible by 6 (since it is divisible by 2 & 3); so there we have 5! = 120 numbers.
Adding all the above counts - 120 + 120 + 120 + 48 + 120 + 120 = 648.
So there are 648 six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place.
