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factorial solving

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by gmataug08 » Wed Nov 26, 2008 11:10 pm
If [(13!)^16 - (13!)^8] / [(13!)^8 + (13!)^4] = a , what is the unit's digit of a/(13!)^4?

a)0
b)1
c)3
d)5
e)9

OA: e
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Re: factorial solving

by srisl11 » Wed Nov 26, 2008 11:50 pm
gmataug08 wrote:If [(13!)^16 - (13!)^8] / [(13!)^8 + (13!)^4] = a , what is the unit's digit of a/(13!)^4?

a)0
b)1
c)3
d)5
e)9

OA: e
I applied a^2 - b^2 = (a-b)(a+b) for the numerator

a= [ (13!^8 + 13!^4)(13!^8 - 13!^4) ]/ [(13!)^8 + (13!)^4]

a = 13!^8 - 13!^4

a = 13!^4 (13!^4 -1)

a/ 13!^4 = 13!^4 -1

The units digit of 13! or 13!^4 will always be 0 (since 13! has 10 in it )

Therefore units digit a/ 13!^4 is 9 (0-1)
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by vishubn » Thu Nov 27, 2008 12:05 am
The units digit of 13! or 13!^4 will always be 0 (since 13! has 10 in it )
NICE ! :)

Vishu
KILL !! DIE !! or BEAT my FEAR !!! de@D END!!
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