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dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Wed May 13, 2009 5:06 pm

The remainder of the division of (1! + 2! + 3! + … + 17!) by 6 is
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5

Source: Beat The GMAT — Problem Solving | Wed May 13, 2009 5:06 pm

cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pm; Thanked: 331 times; Followed by:11 members

by cramya » Wed May 13, 2009 5:22 pm

3

1!+2! = 3

The rest all give a remainder 0 i.e. starting from 3!

sureshbala: Master | Next Rank: 500 Posts; Posts: 319; Joined: Wed Feb 04, 2009 10:32 am; Location: Delhi; Thanked: 84 times; Followed by:9 members

by sureshbala » Thu May 14, 2009 3:55 am

cramya wrote:3

1!+2! = 3

The rest all give a remainder 0 i.e. starting from 3!

Perfect....

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