13 16 - 13 8 13 8 13 4 find the unit digit of a 13 4

a=((13!)^16 -(13!)^8)/((13!)^8+(13!)^4)

=> {(13!)^8 * [(13!)^8 - 1]} / {(13!)^4 * [(13!)^4 + 1]}

=> {(13!)^4 * [((13!)^4)^2 - (1)^2 ] }/ {(13!)^4 + 1 }
using a^2-b^2=(a+b)(a-b)

=> {(13!)^4 [((13!)^4 + 1)((13!)^4 - 1)] } {(13!)^4 + 1 }

=> a= (13!)^4 * [(13!)^4 - 1]

unit digit of a/(13!)^4
=>unit digit of [(13!)^4 - 1]
=> it is clear that the unit digit of (13!)^4 is zero as 13! itself has three zeros
=>So when a number with several zeroes is subtracted by 1 clearly its last digit will be 9(10-1)
=> So unit digit is 9
10 years agoHelpfull: Yes(13) No(1)
Ans: 9
a=((13!)^16 -(13!)^8)/((13!)^8+(13!)^4)
we have to find a/13!^4
=>((13!)^16 -(13!)^8)/13!^4((13!)^8+(13!)^4)
in both numerator and denomenator 13!^8 is common
hence cancel out.
now we have
=>((13!)^8 -1)/((13!)^4+1)
using x^2-y^2=(x+y)(x-y)
x+y will cancel out
we have x-y
=> (13!)^4-1
now we have two parts (13!)^4 and -1
every factorial beyond 4, i.e. 5 have 0 at unit place
and so 0^4 will also produce unit place 0
now we have two element left ac to unit place 0-1
this will produce 9 as unit place in decimal number system
eg: 20-1=19, 30-1=29, 50-1=49
therefore answer is 9 as unit digit.
10 years agoHelpfull: Yes(7) No(1)
pls explain how(elaborate)
-------- unit digit of (13!)^4 is 0
----&----so,unit digit of [(13!)^4-1]= 9 happen i didn't get it.
10 years agoHelpfull: Yes(0) No(0)
13! it contains zero on last digits...and
13!^4-1=like 10-1=9
therefore the ans will be 9
10 years agoHelpfull: Yes(0) No(0)

((13!)^16 -(13!)^8)/((13!)^8+(13!)^4) find the unit digit of a/(13!)^4