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.
This problem is actually much simpler to solve than it seems. We can even generalize a bit and have our product go to 4x, for a function N(x) as the product of the first 4x factorials.
Evaluating N(5) at x=5 for this problem.
Group the terms in pairs:
N(4x) = [1!*2!]*[3!*4!]*[5!*6!]*...*[(4x-1)!*(4x)!]
Factor out the largest term of the even factorial in each pair:
N(4x) = [1!*1!*2]*[3!*3!*4]*[5!*5!*6]*...*[(4x-1)!*(4x-1)!*(4x)]
Now separate the factorials from the independent terms:
N(4x) = [2*4*6*...*(4x)] * [1!*1!]*[3!*3!]*[5!*5!]*...*[(4x-1)!*(4x-1)!]
The right product is a perfect square. For the left product, pull a factor of 2 out of each term:
N(4x) = 2^(2x) * [1*2*3*...*(2x)] * [1!*3!*5!*...*(4x-1)!]^2
Now 2^2x is also a perfect square and the middle product is a factorial, so finally:
N(4x) = (2x)! * [(2^x)*(1!*3!*5!*...*(4x-1)!)]^2
Then we can conclude that removing (2x)! from the product of the first 4x factorials will result in the remaining product being a perfect square.
For the specific problem at hand, x=5 so we are to remove 10! from the product N = (1!)*(2!)*(3!)*(4!)*.....(19!)*(20!) to make the product a perfect square.
Solution
