Factorial Removal Resolution (Posted on 2021-12-01)

 N = (1!)*(2!)*(3!)*(4!)*.....(19!)*(20!).

	Submitted by K Sengupta
Rating: 4.0000 (1 votes)
Solution:	(Hide)
1!*2!*3!*4!......19!*20! = (1!*2!)*(3!*4!)*(5!*6!)*.......(19!*20!) = (1!*2!)*(3!*4*3!)*(5!*6*5!).....(19!*20*19!) = {(1!)*(3!)*(5!).....(19!)}2*(2*4*6*....20) Since the previous expression is a perfect square, it follows that the factorial removal needs to be performed in relation to the expression (2*4*6*.....20) Now, 2*4*6*.....*18*20 = 210*(1*2*3*4*....*9*10) =210*(10!) Since 210 is a perfect square, it follows that removing 10! will render the resultant expression as a perfect square. Also refer to the solution to GENERALIZED VERSION of the problem which has been submitted by Brian Smith in this location. It may be observed that the given problem is equivalent to the Generalised Version for x=5

	Subject	Author	Date
	soln	Steven Lord	2021-12-01 22:14:45
	Solution	Brian Smith	2021-12-01 12:19:00
	solution with a little help from the computer	Charlie	2021-12-01 11:50:44