If n is a positive integer, what is the remainder
When (7^(4n+3))(6^n) is divided by 10?
F. 1
G. 2
H. 4
I. 6
J. 8
When (7^(4n+3))(6^n) is divided by 10?
F. 1
G. 2
H. 4
I. 6
J. 8
7 ^ (4n+3) * 6^ n
the number 7 has a property that when it is raised to a power,
The last digit of the product will repeat after every 4 continous powers
i.e 7 ^ 1 =7
7 ^ 2 =49
7 ^ 3 = 343
7 ^ 4 = ...1
7 ^ 5 =....7 (repetition starts here)
Hence as we put values of n in 7^ (4n+3) like 1,2...
so n =1 will give 7 ^ 7 whose last digit would be same as 7^ 3 (7-3 =4)i.e 3 hence for all +ve values of n 7^ (4n+3) will have last digit as 3
similary 6 ^n will always give last digit 6 as 6 X 6 =36, 36 X 6 =..6 so on
hence upon multiplying 7^ (4n+3) with 6 ^ n
we are multiplying their last digits i.e 3 X 6 =18
hence the product will always end in 8
upon division with 10 it will always give a remainder of 8
hence ans should be E (J here) "8"
