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

Six-digit numbers formed using digits 4, 5, 9 only, divisible by 6.

A number is divisible by 6 if it is divisible by both 2 and 3.

Divisibility by 2: The last digit must be even. Among {4, 5, 9}, only 4 is even. So the units digit must be 4.

Divisibility by 3: The sum of all digits must be divisible by 3.

$$4 \equiv 1 \pmod{3}$$, $$5 \equiv 2 \pmod{3}$$, $$9 \equiv 0 \pmod{3}$$.

With the last digit fixed as 4 (contributing 1 mod 3), we need the sum of the first 5 digits $$\equiv 2 \pmod{3}$$.

Each of the 5 positions can be 4, 5, or 9 independently. Total sequences = $$3^5 = 243$$.

For each position, the digit contributes residues 1, 2, or 0 with equal probability (one digit for each residue class). By symmetry, exactly $$\frac{1}{3}$$ of all sequences have sum $$\equiv 0$$, $$\frac{1}{3}$$ have sum $$\equiv 1$$, and $$\frac{1}{3}$$ have sum $$\equiv 2 \pmod{3}$$.

Number of valid sequences = $$\frac{243}{3} = 81$$.

The answer is 81.