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How to solve this one?

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by vaibhavob » Sun Jan 23, 2011 4:23 am
If ((13!)^16 - (13!)^8)/((13!)^8 + (13!)^4) = a, what is the units digit of a/(13!)^4?

A. 0
B 1
C 3
D 5
E 9

Please suggest how to go about this one? Not able to figure out the trick behind dealing with factorials?
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by maihuna » Sun Jan 23, 2011 4:30 am
See any factorial more than 5 will end in 0, 2*5 = 10 , once there is zero there is always a zero

((13!)^16 - (13!)^8)/((13!)^8 + (13!)^4)
= (13!)^8( ((13!)^8-1))/(((13!)^4(((13!)^4+1)))
= (13!)^4*(!13)^4-1)
Last edited by maihuna on Sun Jan 23, 2011 4:40 am, edited 1 time in total.
Charged up again to beat the beast :)
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by Rahul@gurome » Sun Jan 23, 2011 4:34 am
If ((13!)^16 - (13!)^8)/((13!)^8 + (13!)^4) = a, what is the units digit of a/(13!)^4?
Say 13! = x

Hence, a = (x^16 - x^8)/(x^8 + x^4)
Take x^8 out of bracket in the numerator and x^4 in denominator.

Thus, a = [(x^8)*(x^8 - 1)]/[(x^4)*(x^4 + 1)] = (x^4)*(x^8 - 1)/(x^4 + 1)

Now, (x^8 - 1) = (x^4 + 1)*(x^4 - 1)

Hence, a = (x^4)(x^4 - 1)

Thus, a/(13!)^4 = a/(x^4) = (x^4 - 1) = (13!)^4 - 1

Now the unit's digit of 13! is 0.
Thus, unit's digit of (13!)^4 is also 0.
Thus, unit's digit of [(13!)^4 - 1] is 9.

The correct answer is E.
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by vaibhavob » Sun Jan 23, 2011 9:27 pm
Thanks guys for the explanation.Now, I am clear with how to deal with factorials.

Regards,

Vaibhav
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