October 2017 Problem

What factor should be crossed out of the product

 

1! х 2! х 3! х ............х 16! ,

to make the remaining expression a perfect square ?

 

Dear friends and visitors of the site.

Thank you for your solutions to the October Problem.

The first to send a correct solution was Ruining and Asker.

Our congrats to the winners!

Here is the correct solution:

We remark that

1! х 2! х 3! х 4! х.......х 15! x16 ! = (1! х 2!) х (3! х 4!) х..........х (15! х 16!) = = (1! х 1! х 2) х (3! х 3! х 4) х (5! х 5! х 6) х...........х (13! х 13! х 14) х (15! х 15! х 16) = = (1!)^2 х (3!)^2 х (5!)^2 х............х (15!)^2 х (2 х 4 х 6 х 8 х...........х 14 х 16) = = (1!)^2 х (3!)^2 х (5!)^2 х.............х (19!)^2 х (2 х (2 х 2) х (3 х 2) х..............х (8 х 2)) = = (1! х 3! х............х 15!)^2 х 2^8 х (1 х 2 х 3 х...............х 8) = (1! х 3! х..............х 15!)^2 (2^4)^2 х 8!

We see that the first two factors are squares and that crossing out 8! gives us a perfect square.

Thus the answer is  8!

Thank you again and welcome to the November Problem!

Serge Hazanov