Question

The remainder obtained when $(2+5^2+6^3+{11}^4+2^5+5^6+6^7+{11}^8+\dots \dots 2^{97}+5^{98}+6^{99}+{11}^{100})$ is divided by 5 is

• A. 0
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A

0

The unit digit of $2,5^2,6^3,{11}^4$ are 2,5,6,1. For the next four consecutive terms also have same unit digits and so on. The sum of unit digits is 14. This will happen for 25 times, so unit digit is $14\times 25=350$. So the unit digit is 0. Hence, the remainder is 0.