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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 $1395t+2015$

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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 $1$.

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