The remainder when $(15^{23} + 23^{23})$ is divided by 19, is:

Finding the remainder when $(15^{23} + 23^{23})$ is divided by 19.

When dividing by 19, we work with remainders:

$15 \equiv 15 \pmod{19}$

$23 = 19 + 4$

$23 \equiv 4 \pmod{19}$

The problem becomes finding the remainder of $(15^{23} + 4^{23})$ when divided by 19.

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By Fermat's Little Theorem, for prime $p = 19$:

$a^{18} \equiv 1 \pmod{19}$

Since $23 = 18 + 5$:

$15^{23} = 15^{18} \times 15^5$

$15^{23} \equiv 1 \times 15^5 \pmod{19}$

$4^{23} = 4^{18} \times 4^5$

$4^{23} \equiv 1 \times 4^5 \pmod{19}$

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Calculating $15^5 \pmod{19}$:

$15^2 = 225$

$225 = 11(19) + 16$

$15^2 \equiv 16 \pmod{19}$

$15^4 = (15^2)^2$

$15^4 \equiv 16^2 = 256 \pmod{19}$

$256 = 13(19) + 9$

$15^4 \equiv 9 \pmod{19}$

$15^5 = 15^4 \times 15$

$15^5 \equiv 9 \times 15 = 135 \pmod{19}$

$135 = 7(19) + 2$

$15^5 \equiv 2 \pmod{19}$

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Calculating $4^5 \pmod{19}$:

$4^2 = 16 \pmod{19}$

$4^4 = (4^2)^2$

$4^4 \equiv 16^2 = 256 \pmod{19}$

$256 = 13(19) + 9$

$4^4 \equiv 9 \pmod{19}$

$4^5 = 4^4 \times 4$

$4^5 \equiv 9 \times 4 = 36 \pmod{19}$

$36 = 1(19) + 17$

$4^5 \equiv 17 \pmod{19}$

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$(15^{23} + 4^{23}) \equiv 15^5 + 4^5 \pmod{19}$

$(15^{23} + 4^{23}) \equiv 2 + 17 \pmod{19}$

$(15^{23} + 4^{23}) \equiv 19 \pmod{19}$

$(15^{23} + 4^{23}) \equiv 0 \pmod{19}$

Therefore, the remainder is $0$.

The answer is option 4.