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# Is ! even? (1) E=Q (2) Q=2

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VP
Joined: 13 Jun 2004
Posts: 1122
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 31 [0], given: 0

D for sure

the only way to get a odd factorial is with 1!=1
and the only "special" factorial to consider would be 0! = 0

In the first statement we had to get E=Q
0! -> (0+0)! = 0 -> EVEN
1! -> (1!+1!)! = 2! -> EVEN

so statement 1 is sufficient

In the second statement it's even anyway (any factorial number superior to 1 will be even as it will be multiply by 2)

so statement 2 is sufficient

Manager
Joined: 28 Aug 2004
Posts: 205
Followers: 1

Kudos [?]: 1 [0], given: 0

Caspace wrote:
unless i am crazy you couldn't have factorials without using integers right?

Based on that assumption i throw in for D too.

You could unless the word 'factorial' is associated with the word 'integer' or with an integer number, which is the case in most problems.
Senior Manager
Joined: 19 Nov 2004
Posts: 284
Location: Germany
Followers: 1

Kudos [?]: 13 [0], given: 0

I think the factorial sign outside the bracket makes the difference.

I'll go with D

good discussion folks!

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