Suppose that 6-digit numbers are formed using each of the digits 1,2,3,7,8,9 exactly once. I want the count of such 6-digit numbers that are divisible by 6 but not divisible by 9. I understand that the formed number should contain atleast one 2 and atmost one 3 in its prime factorization but don't know how to find count of such numbers. Any hint in this direction or other solution method will be very helpful.
Count of 6-digit numbers divisible by 6 but not by 9
divisibilitynumber theory
Best Answer
Firstly, we know that any number of the combination of digits 1,2,3,7,8,9 with each number used exactly once is not divisible by 9, since $ 1 + 2 + 3 + 7 + 8 + 9 = 30 $ which is not divisible by 9.
Secondly, to let the number be divisible by 6, the last digit must be even. Therefore the last digit can only be 2 or 8.
Since there is no repeating digit, no repetition will need to be considered and the answer is $ 2 \times 5! = 240 $ .