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)
How to solve this one?
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Source: Beat The GMAT — Data Sufficiency |
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Rahul@gurome
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Say 13! = xIf ((13!)^16 - (13!)^8)/((13!)^8 + (13!)^4) = a, what is the units digit of a/(13!)^4?
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.
Rahul Lakhani
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Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
