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26 Dec 2011, 11:29
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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11. If (13!)^16 - (13!)^8 / (13!)^8 - (13!)^4 = a, then what is the units digit of a/(13!)^4
(A) 0
(B) 1
(C) 3
(D) 5
(E) 9
OA is E. Can someone please help? This is how I am trying to solve this
Numerator can be broken down into
[(13!)^8 (13!)^8] - [(13!)^4 (13!)^4] => [(13!)^8]^2 - [(13!)^4]^2
[(13!)^8 + (13!)^4] [(13!)^8 - (13!)^4] --------------------------- (x^2-y^2) = (x+y) (x-y)
[(13!)^8 + (13!)^4] [(13!)^8 - (13!)^4] / [(13!)^8 + (13!)^4] => [(13!)^8 - (13!)^4]
(13!)^4 [(13!)^4-1] = a
[(13!)^4-1] = a/(13!)^4
I am stuck after this. Can someone please help?
(A) 0
(B) 1
(C) 3
(D) 5
(E) 9
OA is E. Can someone please help? This is how I am trying to solve this
Numerator can be broken down into
[(13!)^8 (13!)^8] - [(13!)^4 (13!)^4] => [(13!)^8]^2 - [(13!)^4]^2
[(13!)^8 + (13!)^4] [(13!)^8 - (13!)^4] --------------------------- (x^2-y^2) = (x+y) (x-y)
[(13!)^8 + (13!)^4] [(13!)^8 - (13!)^4] / [(13!)^8 + (13!)^4] => [(13!)^8 - (13!)^4]
(13!)^4 [(13!)^4-1] = a
[(13!)^4-1] = a/(13!)^4
I am stuck after this. Can someone please help?
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26 Dec 2011, 17:59
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@enigma123 - Is the denominator (13!)^8 - (13!)^4 as written in the question or (13!)^8 + (13!)^4 as written in your steps towards a solution? Based on the OA, I assume it's "+" not "-" but I wanted to confirm.
Thanks!
Bhavin
Thanks!
Bhavin
26 Dec 2011, 21:26
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enigma as per your calculation we get
[(13!)^4-1] = a/(13!)^4 as the final solution,
Now 13!=13*12*11*10*9*8*7*6*5*4*3*2*1= x * 10^2
we can see that there are 2 10's in the answer of 13! (10 and 5*2) hence the answer will always contain 2 zeros. now,
(x * 10^2)^4=>more zeros in the end.
Hence if you subtract 1 from (13!)^4 you will always get 9 in the units digit.
I hope this helps!
[(13!)^4-1] = a/(13!)^4 as the final solution,
Now 13!=13*12*11*10*9*8*7*6*5*4*3*2*1= x * 10^2
we can see that there are 2 10's in the answer of 13! (10 and 5*2) hence the answer will always contain 2 zeros. now,
(x * 10^2)^4=>more zeros in the end.
Hence if you subtract 1 from (13!)^4 you will always get 9 in the units digit.
I hope this helps!
