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 ?

Answer Submission Is Not Available

Form is valid through October 2017

Solution

 

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