Remainder

$5^{2}\equiv{2}\pmod{23}$

$5^{4020}\equiv{2^{2010}}\pmod{23}$

Hence $5^{4020}$ and $2^{2010}$ both leave the same reminder when divided by $23$ .

So asked thing is reduced to finding $2^{2014}+7\cdot2^{2010}\pmod{23}$

$2^{2014}+7.2^{2010}$

= $2^{2010}(2^{4}+7)$

= $2^{2010}.23$

Hence.the expression is divisible by $23$ and leaves a remainder of $0$ when divided by $23$ .

By Fermat's little theorem , as 23 is a prime,

$2^{22} \equiv 1 \pmod{23}\\ \therefore 2^{22\times 91}=2^{2002} \equiv 1 \pmod{23} \\ \therefore 2^{2014} \equiv 2^{12} \pmod{23}$

---

Similarly doing for the second term, we have

$5^{4004}\equiv 1 \pmod{23} \\ \therefore 5^{4020} \equiv 5^{16} \pmod{23}$

---

Moreover, we have

$5^2 \equiv 25\equiv 2\pmod{23} \\ \implies 5^{16} \equiv 2^8 \pmod{23}$

---

So asked thing is reduced to finding $2^{12} + 7\times 2^{8} \pmod{23}$

This is simple by following way, $2^{12} + 7\times 2^8\equiv 2^8 (2^4+7) \equiv 2^8(23) \equiv \boxed{0}\pmod{23}$