$\begin{array}{1 1} 216 \\ 240 \\ 720 \\ 96 \end{array}$

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Since the numbers are divisible by 5, in last place (unit's place)

we can have either 5 or 0

Fixing 0 in the last place we can have $5!$ =120 numbers.

Similarly,by fixing 5 at the last place we can have 120 numbers.

But out of these 120 numbers we have to eliminate

the numbers starting with 0.

Because these numbers will be 5 digit numbers.

$i.e.,$ Fixing 0 at the first place and 5 at the last place,

we can have $4!=24$ numbers which are 5 digit numbers.

$\therefore$ The required 6 digit numbers = $120+120-24=216$

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