GeminiHeat wrote:
If \(a=\frac{13!^1^6-13!^8}{13!^8+13!^4}\) what is the unit digit of \(\frac{a}{13!^4}\)?
A. 0
B. 1
C. 9
D. 4
E. 6
\(a = \frac{13!^{16} - 13!^8}{13!^8 + 13!^4}\)
Let, \(13! = x\)
\(a = \frac{(x^8 - x^4)(x^8 + x^4)}{(x^8 + x^4)}\)
\(a = x^8 - x^4\)
Now, \(\frac{a}{x^2} = x^2 - 1\)
Since we have \(13!\), we will have a \(10\) in it which makes it's unit digit as \(0\)
Therefore, \((13!)^2\) will have units digit as \(0\)
Our Answer would be \(.....0 - 1 = 9\)
Hence, option C