Unit's Digit?

Method 1 -- by Binomial Expansion

$\begin{aligned} 13^{17} + 17^{13} & \equiv (10+3)^{17} + (10+7)^{13} \text{ (mod 10)} \\ & \equiv 3^{17} + 7^{13} \text{ (mod 10)} \\ & \equiv 3(9^8) + 7(49^6) \text{ (mod 10)} \\ & \equiv 3(10-1)^8 + 7(50-1)^6 \text{ (mod 10)} \\ & \equiv 3(-1)^8 + 7(-1)^6 \text{ (mod 10)} \\ & \equiv 3 + 7 \text{ (mod 10)} \\ & \equiv \boxed{0} \text{ (mod 10)} \end{aligned}$

Method 2 -- by Euler's Theorem : Since 13, 17 and 10 are coprime integers, we can use Euler's theorem. We note that Euler's totient function $\phi(10) = 10 \times \dfrac 12 \times \dfrac 45 = 4$ .

$\begin{aligned} 13^{17} + 17^{13} & \equiv 13^{17 \text{ mod }\phi(10)} + 17^{13 \text{ mod }\phi(10)} \text{ (mod 10)} \\ & \equiv 13^{17 \text{ mod }4} + 17^{13 \text{ mod }4} \text{ (mod 10)} \\ & \equiv 13^1 + 17^1 \text{ (mod 10)} \\ & \equiv \boxed{0} \text{ (mod 10)} \end{aligned}$

I used the cyclic app are ancestry of the last number: for 13^n it has a as last when n = 1 mod 4. 17^n has a 7 for n = 1 mod 4, so 3 + 7 =10. Hence the unit digit is 0.

$(13^{17}+17^{13})\mod 10= (3^{17}+7^{13 })\mod 10 = 3 + 7= 10$ . Therefore, the units digit is $0$ .