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Number of zeroes at the end of the expression (2!)^2!+4!^4!+8!^8!+9!^9!+10!^10!+11!^11!

+3 votes
asked Nov 4, 2013 in TCS by Campusgate

2 Answers

+1 vote
 
Best answer
To form a zero we need one 2 and one 5.  In general any factorial contains more 2's than 5's. So limiting factor is 5's.  We will find how many 5's each factorial has.
2! and 4! does not contain any zeros. 8!, 9! contains 1 and 10! and 11! contains 2 each.
So total zeroes are 1 x 8! + 1 x 9! + 2 x 10! + 2 x 11! = 8!+9!+2(10!)+2(11!)

For more details on this type of question read this article: http://www.campusgate.co.in/2011/10/finding-maximum-power-of-number-divide.html
answered Nov 6, 2013 by Campusgate
0 votes
8!+9!+2(10!)+2(11!)
answered Nov 5, 2013 by Anand
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