Question

The total number of six digit numbers, formed using the digits 4,5,9 only and divisible by 6 , is __

Hint:

To check divisibility by 6, always verify divisibility by both 2 (even last digit) and 3 (sum of digits divisible by 3).

Answer 1

Correct Answer: 81

The number must be divisible by \(6\), so it must be divisible by both \(2\) and \(3\). For divisibility by \(2\), the last digit must be \(4\). For divisibility by \(3\), the sum of the digits must be divisible by \(3\).
Let us consider different cases based on the distribution of digits:
Case 1: All digits are the same
For \(44444\), there is only \(1\) number.
Case 2: Two distinct digits
\((4, 5)\): Numbers of the form \(44455\):
\[\frac{5!}{3!2!} = 10.\]
\((4, 9)\): Numbers of the form \(44499\):
\[\frac{5!}{3!2!} = 10.\]
Case 3: Three distinct digits
For \(4, 5, 9\), let us consider the permutations:
Digits: \(4, 5, 9, 4, 4\):
\[\frac{5!}{3!} = 20.\]
Digits: \(4, 5, 9, 5, 5\):
\[\frac{5!}{3!2!} = 5.\]
Digits: \(4, 5, 9, 9, 9\):
\[\frac{5!}{3!2!} = 5.\]
Digits: \(4, 5, 9, 4, 5\):
\[\frac{5!}{2!2!1!} = 30.\]
Total
\[1 + 10 + 10 + 20 + 5 + 5 + 30 = 81.\]
Conclusion
The total number of such numbers is:
\[\boxed{81}.\]