Difficult Math Question #23 - Algebra

[800guy]

OA coming after few people have answerd--see attachment

[gamemaster]

x = 13!

a = x^16 - x^8 / x^8 + x ^ 4 =

(x^8+x^4)(x^8-x^4)/(x^8+x^4) =

x^8 - x^4

a/x^4 =

x^4 - 1

digit number of 13! must be 0 because its like (1*2*3...*11*12*13) * 10

so the final answer would be 1

[sanjeevs50]

(E) is the anwer

0 of unit digit - 1 = 9

[shail_8206]

Let (13!)^4 = x
(x^4 - x^2)/ (x^2 - x) = a
by solving further we get:
x^3 + x^2 = ax
or a/x = x + 1
a/(13! ^ 4 ) = (13!^ 4) + 1

as 13! ends with 0, so is 13!^ 4; so (13!^ 4) + 1 ends with 1.
Unit's digit is 1.

[gamemaster]

the answer is 9

i had a typo...

[800guy]

here's the oa, look at the attachment