Can you order the digits $1, 2, 3, 4, 5$ and $6$ to make a number which is divisible by $6$ ...
... so that when the final or last digit is removed it becomes a $5$-figure number divisible by $5$?
And when the final digit is removed again it becomes a $4$-figure number divisible by $4$?
And when the final digit is removed again it becomes a $3$-figure number divisible by $3$?
And when the final digit is removed again it becomes a $2$-figure number divisible by $2$, then finally a $1$-figure number divisible by $1$?