If $$\sum\limits_{r = 1}^{10} {r!({r^3} + 6{r^2} + 2r + 5) = \alpha (11!)} $$, then the value of $$\alpha$$ is equal to ___________.

The number of times the digit 3 will be written when listing the integers from 1 to 1000 is :

The missing value in the following figure is

The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is _____________.