Find sum of the digits of $1395t+2015$

Assume $$S(x)$$ be equal to sum of digits of the integer $$x$$. for example $$S(123)=1+2+3=6$$. suppose $$t$$ be smallest natural number so that $$S(t)=1394^{1394}$$. then what is the value of $$S(1395t+2015)$$?

$$1)26 \quad\quad2)35\quad\quad3)134\quad\quad4)143\quad\quad5)\text{None}$$

To solve this problem first of all I should find the value of $$t$$. because $$t$$ is smallest natural number with digits sums to $$1394^{1394}$$, therefore we should use the digit $$9$$ as much as possible to reduce the number of digits. $$1394^{1394}\equiv(-1)^{1394}\equiv1\pmod{9}$$so $$t=1\underbrace{999\cdots9}_{\tfrac{1394^{1394}-1}9}$$ But I don't know how to find sum of the digits of $$1395t+2015$$

You have made a great start. To continue, let

$$n = \frac{1394^{1394}-1}{9} \tag{1}\label{eq1A}$$

This then gives

$$t = 2 \times 10^{n} - 1 \tag{2}\label{eq2A}$$

Thus,

\begin{aligned} 1395t + 2015 & = 1395(2 \times 10^{n} - 1) + 2015 \\ & = 2790 \times 10^{n} - 1395 + 2015 \\ & = 2790 \times 10^{n} + 620 \end{aligned}\tag{3}\label{eq3A}

Note $$n$$ is much larger than $$3$$, so there's no overlap between the the highest non-zero digits of $$279$$ and the final digits of $$620$$, with all of the digits in between being $$0$$. This means that

$$S(1395t + 2015) = 2 + 7 + 9 + 6 + 2 = 26 \tag{4}\label{eq4A}$$

i.e., the listed choice $$1$$.