Find the ones/units digit

Let $A$ be the number given.

$\begin{aligned} A &= 17^{182}\times8^{433}+23^{79}\times11^{94} \\ & =(17^{4})^{45}\times17^{2}\times(2^{10})^{129}\times2^{9} + (23^{4})^{19}\times23^{3}\times11^{94} \\ & = (83521)^{45}\times289\times(1024)^{129}\times(2)^{9} + (279841)^{19}\times(23)^{3}\times(11)^{94} \\ & = 1\times9\times4\times2 + 1\times7\times1 \\ & = 2+ 7 = \boxed{9} \end{aligned}$

Relevant wiki: Euler's Theorem

Let $N$ be the number given. We need to find $N \bmod 10$ .

$\begin{aligned} N & \equiv 17^{\color{#3D99F6}182 \bmod \phi(10)} \left(2^3\right)^{433} + 23^{\color{#3D99F6}79 \bmod \phi(10)}(10+1)^{96} \text{ (mod 10)} &\small \color{#3D99F6} \text{Since }\gcd (17,10) = \gcd(23, 10) = 1 \text{, Euler's theorem applies.} \\ & \equiv 17^{\color{#3D99F6}182 \bmod 4} \left(2^{4 \times 324+3}\right) + 23^{\color{#3D99F6}79 \bmod 4}(1) \text{ (mod 10)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(10) = 4 \\ & \equiv 17^{\color{#3D99F6}2} \left({\color{#D61F06}16^{324}}\times 8\right) + 23^{\color{#3D99F6}3} \text{ (mod 10)} & \small \color{#D61F06} \text{Power of integer ends with 6 always ends with 6.} \\ & \equiv 7^2\times {\color{#D61F06}6} \times 8 + 3^3 \text{ (mod 10)} \\ & \equiv 9 \times 8 + 7 \text{ (mod 10)} \\ & \equiv 2+7 \text{ (mod 10)} \\ & \equiv \boxed{9} \text{ (mod 10)} \end{aligned}$

Considering the first term: $17^{182}$

$7^1=7,7^2=49,7^3=343,7^4=2401,7^5=16807$ observed that the units digit repeat in every cycle of $4$ . So we divide the exponent by $4$ then find the remainder.

$\dfrac{182}{4}=45 \dfrac{2}{4}$ : the remainder is $2$ so the last digit is $9$ .

Considering the second term: $8^{433}$

$8^1=8, 8^2=64, 8^3=512, 8^4=4096, 8^5=32768$ observed that the units digit repeat in every cycle of 4. So we divide the exponent by $4$ then find the remainder.

$\dfrac{433}{4}=108 \dfrac{1}{4}$ : the remainder is $1$ so the last digit is $8$

Considering the third term: $23^{79}$

$3^1=3,3^2=9,3^3=27,3^4=81,3^5=343$ observed that the units digit repeat in every cycle of 4. So we divide the exponent by $4$ then find the remainder.

$\dfrac{79}{4}=19 \dfrac{3}{4}$ : the remainder is $3$ so the last digit is $7$ .

Considering the fourth term: $11^{94}$

$1^1=1,1^2=1,1^3=1$ observed that the last digit is always $1$ so the last digit of the fourth term is $1$ .

The last digit of the first two terms is $2$ $(9 \times 8=72)$ .

The last digit of the last two terms is $7$ $(7 \times 1 = 7)$ .

Adding the two gives

$2+7=\boxed{9}$