Show that for all integer values of ,
. . is divisible by
Hello, Galactus!
Of course, you meant: 5 is a factor of
Then your solution is correct.
The "Quickie" solution goes like this:
We know that the product of consecutive integers is divisible by
. . and the product of four consecutive integers is divisible by
We have: .
. . and: .
. . then is divisible by
. . then is divisible by
. . then is divisible by
Therefore, is divisible by