That Is Close 5

Let $S=15^{23} +23^{23}$

We individually find the remainder after dividing by $3$ and $19$ . because $57=19 \times 3$

$15^{23} = 3^{23}.5^{23} \equiv 0 \pmod 3$

$23^{23}=(24-1)^{23} \equiv (-1)^{23} \equiv -1 \equiv 2 \pmod 3$ .So $15^{23} + 23^{23} \equiv 2 \pmod 3$

Now $15^{23} + 23^{23} =(19-4)^{23} + (19+4)^{23} =19k_1 +(-4)^{23} + 19k_2 +4^{23}=19k \equiv 0 \pmod{19}$

So $S \equiv 2 \pmod{3} \\ S \equiv 0 \pmod{19}$ .

The smallest number that satisfies the above condition is $R=38$

So $38 \equiv 3 \pmod{5}$

We need to find $R = 15^{23}+23^{23} \text{ mod }57$ and then $R \text{ mod }5$ .

Since 15 and 57 are not coprime integers, we have to consider the prime factors of 57, 3 and 19, separately.

$\begin{aligned} 15^{23} & \equiv 0 \text{ (mod 3)} \end{aligned}$

$\begin{aligned} \implies 15^{23} & \equiv 3{\color{#3D99F6}n} \text{ (mod 57)} & \small \color{#3D99F6} \text{where }n \text{ is a positive integer.} \end{aligned}$

$\begin{aligned} 15^{23} & \equiv 15^{23 \text{ mod }\color{#3D99F6}\phi(19)} \text{ (mod 19)} & \small \color{#3D99F6} \text{Since }\gcd(15,19) = 1 \text{, Euler's theorem applies.} \\ & \equiv 15^{23 \text{ mod }\color{#3D99F6}18} \text{ (mod 19)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(19) = 18 \\ & \equiv 15^5 \text{ (mod 19)} \\ & \equiv (19-4)^5 \text{ (mod 19)} \\ & \equiv -4^5 \text{ (mod 19)} \\ & \equiv -(16)(16)4 \text{ (mod 19)} \\ & \equiv -(-3)(-3)4 \text{ (mod 19)} \\ & \equiv -36 \text{ (mod 19)} \\ & \equiv 2 \text{ (mod 19)} \end{aligned}$

$\begin{aligned} \implies 3n & \equiv 2 \text{ (mod 19)} \\ 21 & \equiv 2 \text{ (mod 19)} \\ \implies n & = 7 \\ \implies 15^{23} & \equiv 21 \text{ (mod 57)} \end{aligned}$

Although 23 and 57 are coprime integers, $\phi(57) = 36 > 23$ is unsuitable to solve for $23^{23} \text{ mod }57$ . We again consider the prime factors, 3 and 19 separately, which are both coprime integers with 23 and Euler's theorem applies.

$\begin{aligned} 23^{23} & \equiv 23^{23 \text{ mod }\color{#3D99F6}\phi(3)} \text{ (mod 3)} \\ & \equiv 23^{23 \text{ mod }2} \text{ (mod 3)} \\ & \equiv 23^1 \text{ (mod 3)} \\ & \equiv 2 \text{ (mod 3)} \end{aligned}$

$\begin{aligned} \implies 23^{23} & \equiv 3{\color{#3D99F6}p} + 2 \text{ (mod 57)} & \small \color{#3D99F6} \text{where }p \text{ is a positive integer.} \end{aligned}$

$\begin{aligned} 23^{23} & \equiv 23^{23 \text{ mod }\color{#3D99F6}\phi(19)} \text{ (mod 19)} \\ & \equiv 23^{23 \text{ mod }18} \text{ (mod 19)} \\ & \equiv 23^5 \text{ (mod 19)} \\ & \equiv (19+4)^5 \text{ (mod 19)} \\ & \equiv 4^5 \text{ (mod 19)} \\ & \equiv (16)(16)(4) \text{ (mod 19)} \\ & \equiv (-3)(-3)(4) \text{ (mod 19)} \\ & \equiv 36 \text{ (mod 19)} \\ & \equiv 17 \text{ (mod 19)} \end{aligned}$

$\begin{aligned} \implies 3p + 2 & \equiv 17 \text{ (mod 19)} \\ 3p & \equiv 15 \text{ (mod 19)} \\ p & \equiv 5 \text{ (mod 19)} \\ 24 & \equiv 5 \text{ (mod 19)} \\ \implies p = 24 \\ \implies 3(24) + 2 & \equiv 74 \text{ (mod 57)} \\ & \equiv 17 \text{ (mod 57)} \end{aligned}$

Therefore, $15^{23}+23^{23} \equiv 21 + 17 \equiv 38 \text{ (mod 57)}$ $\implies R = 38$ and $R \text{ mod 5} = 38 \text{ mod 5} = \boxed{3}$ .