gmat6nplus1 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
\(a= \frac{13!^1^6-13!^8}{13!^8+13!^4}\)
\(=\frac{13!^8(13!^8-1)}{(13!^4(13!^4+1)}\)
\(=\frac{13!^4(13!^4+1)(13!^4-1)}{(13!^4+1)}\)
\(=13!^4(13!^4-1)\)
So, \(\frac{a}{13!^4} = (13!^4-1)\)
\(13!^4\) has unit's digit 0
so \((13!^4-1)\) has unit's digit = 10-1 =9
Answer C
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