GATE CE 2017 Set 1 — Question 63

To find the last digit of the sum, we need to find the last digit of each term and then sum their last digits. The last digit of $2171$ is $1$. Any positive integer power of a number ending in $1$ will also end in $1$.
So, the last digit of $(2171)^7$ is $1$. The last digit of $2172$ is $2$. We look at the cyclicity of the last digits of powers of $2$:
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16 \rightarrow 6$
$2^5 = 32 \rightarrow 2$
The cycle of last digits for powers of $2$ is $(2, 4, 8, 6)$, with a length of $4$.
To find the last digit of $2^9$, we divide the exponent $9$ by the cycle length $4$: $9 \div 4 = 2$ with a remainder of $1$.
The last digit is the $1^{st}$ digit in the cycle, which is $2$.
So, the last digit of $(2172)^9$ is $2$. The last digit of $2173$ is $3$. We look at the cyclicity of the last digits of powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27 \rightarrow 7$
$3^4 = 81 \rightarrow 1$
$3^5 = 243 \rightarrow 3$
The cycle of last digits for powers of $3$ is $(3, 9, 7, 1)$, with a length of $4$.
To find the last digit of $3^{11}$, we divide the exponent $11$ by the cycle length $4$: $11 \div 4 = 2$ with a remainder of $3$.
The last digit is the $3^{rd}$ digit in the cycle, which is $7$.
So, the last digit of $(2173)^{11}$ is $7$. The last digit of $2174$ is $4$. We look at the cyclicity of the last digits of powers of $4$:
$4^1 = 4$
$4^2 = 16 \rightarrow 6$
$4^3 = 64 \rightarrow 4$
The cycle of last digits for powers of $4$ is $(4, 6)$, with a length of $2$.
For odd exponents, the last digit is $4$. For even exponents, the last digit is $6$.
Since the exponent $13$ is an odd number, the last digit is $4$.
So, the last digit of $(2174)^{13}$ is $4$.Now, we sum the last digits we found:
Last digit of sum = Last digit of $(1 + 2 + 7 + 4)$
Last digit of sum = Last digit of $(14)$
The last digit of $14$ is $4$.Therefore, the last digit of the given expression is $4$.The final answer is $\boxed{B}$