Sum of last three digits of the number `N=7^(100)-3^(100)` is.

Options

A. 2000
B. 4000
C. 6000
D. 8000

Answer

d

Explanation

`N=7^(100) - 3^(100) = (7^(2))^(50) - (3^(2))^(50)`
`= (50-1)^(50) - (10-1)^(50)`
`=[(50)^(50) -""^(50)C_(1) (50)^(49) + ""^(50)C_(2) (50)^(48) - ""^(50)C_(3)`
`(50)^(47)+...+ ""^(50)C_(48) (50)^(2) - ""(50)C_(49) (50)+1]`
`-[10^(50)-""^(50)C_(1) *10^(49) + ""^(50)C_(2) (10)^(48) - ""^(50)C_(3) (10)^(47) + ...+ ""^(50)C_(45) (10)^(2) - ""^(50)C_(49)(10)+1]`
`=[10^(4)m-""^(50)C_(47) (50)^(3)+ ""^(50)C_(48)(50)^(2) - ""^(50)C_(49) (50)+]`
`-[10^(4)n- ""^(50)C_(47)(10)^(3) + ""^(50)C_(48)(10)^(2) - ""^(50)C_(49)(10)^(2) - ""^(50)C_(49) (10) + 1]`
when m and n are interger
`= 10^(4)m-""^(50)C_(47) [(50)^(3) - (10)^(3)]+""^(50)C_(2) [(50)^(2)- (10)^(2)] - ""^(50)C_(1)[(50)- (10)]`
when m and n are integer
`= 10^(4)p- 124xx196xx10^(5)+ 294xx10^(4)- 2000 = 10^(4) q= 2000`
when q is an integer .
`= 10^(4) q- 10^(4) - 2000 = 10^(4)(q-1) + 8000`