N = (1!)*(2!)*(3!)*(4!)*.....(19!)*(20!).
Precisely 1 of the 20 factorials needs to be removed from N to make it a perfect square.
What factorial needs to be removed? Provide adequate reasoning for your answer.
prod {1! 2! ... 20!)
= prod {2 [3! 3! 4] [5! 5! 6] ... [19! 19! 20] }
= prod { (3!^2 5!^2 ... 19!^2) (2 4 6 8 ... 20) }
= prod { 3! 5! ... 19!}^2 (4 9 16) (2 6 8 10 12 14 18 20)
= prod { 3! 5! ... 19!}^2 (2^2 3^2 4^2) (2 2 3 2 2 2 2 5 2 2 3
2 7 2 3 3 2 2 5)
= prod {3! 5! ... 19!}^2 24^2 2^12 3^4 5^2 7
= {prod(3! 5! ... 19!) 24 2^6 3^2 5}^2 7
So, since the product is a square times seven, we must leave out seven.
How? One of the factorials must be seven times a perfect square. We wish to leave this one out and we will still have a perfect square: if (ab)^2 is a perfect square, a^2 is a perfect square. Which one?
The number must be between 7 and 13 to have exactly one seven in its factorial. There also must be 0, 2 or 4 fives in the number's factorial to be a square, so it must be less than 5, between 10 and 14 inclusive, or 20. So, together with the previous constraint, the number must be between 10 and 13 inclusive. Since a single eleven is in the factorials of 11, 12 and 13, 10! is our only possibility, and 10! works:
10! = 2 3 4 5 6 8 9 10 = (2 2 2 2 2 2 2 2 3 3 3 3 5 5) 7 = 720^2 7
So, leave out 10!
(Of course, Brian's general solution is way cooler)
Edited on December 2, 2021, 4:10 pm
soln
