If [(13!)^16 - (13!)^8] / [(13!)^8 + (13!)^4] = a , what is the unit's digit of a/(13!)^4?
a)0
b)1
c)3
d)5
e)9
OA: e
a)0
b)1
c)3
d)5
e)9
OA: e
I applied a^2 - b^2 = (a-b)(a+b) for the numeratorgmataug08 wrote:If [(13!)^16 - (13!)^8] / [(13!)^8 + (13!)^4] = a , what is the unit's digit of a/(13!)^4?
a)0
b)1
c)3
d)5
e)9
OA: e
a= [ (13!^8 + 13!^4)(13!^8 - 13!^4) ]/ [(13!)^8 + (13!)^4]
a = 13!^8 - 13!^4
a = 13!^4 (13!^4 -1)
a/ 13!^4 = 13!^4 -1
The units digit of 13! or 13!^4 will always be 0 (since 13! has 10 in it )
Therefore units digit a/ 13!^4 is 9 (0-1)
