Finding the last digits of two towers

Note that $\gcd(7,10) = \gcd(3,10)=1$ , we can apply the Euler's theorem in both cases. The Euler's totient function $\phi(10) = 10 \times \times \dfrac 12 \times \dfrac 45 = 4$ .

$\large \begin{aligned} 7^{5^3 \bmod \phi(10)} \equiv 7^{5^3 \bmod 4} \equiv 7^{(4+1)^3 \bmod 4} \equiv 7^{1^3} \equiv 7 \text{ (mod 10)} \\ 3^{5^7 \bmod \phi(10)} \equiv 3^{5^7 \bmod 4} \equiv 3^{(4+1)^7 \bmod 4} \equiv 3^{1^7} \equiv 3 \text{ (mod 10)} \end{aligned}$

Therefore the answer is $7 \times 3 = \boxed{21}$ .

$\large 7^{5^{3}} \normalsize \equiv 7 \pmod{10}$

$\large 3^{5^{7}} \normalsize \equiv 3 \pmod{10}$

$3 \times 7 = \boxed{21}$