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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