TITLE: Find sum of the digits of $1395t+2015$
QUESTION [6 upvotes]: 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 $1395t+2015$

REPLY [3 votes]: 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,
$$\begin{equation}\begin{aligned}
1395t + 2015 & = 1395(2 \times 10^{n} - 1) + 2015 \\
& = 2790 \times 10^{n} - 1395 + 2015 \\
& = 2790 \times 10^{n} + 620
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
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 $1$.