Question

Let $N$ be the smallest positive integer such that $N+2N+3N+\dots \dots +9N$ is a number all whose digits are equal. What is the sum of the digits of $N$?

Answer 1

$N$ is an integer.
$P=N+2N+3N+\dots \dots +9N$
$\Rightarrow P=(1+2+...+9)N=45N$
If $N$ is an even integer, then $P=45N$ will end in $0$.
This is not possible as all the digits are same.
If $N$ is an odd integer, then $P=45N$ will end in $5$.
$\therefore P=45N=\underset{\text{a times}}{\underset{}{555\dots \dots 5}}$
Also as $P=45N$ is a multiple of $9$.
$5$ must appear atleast $9$ times.
$\therefore N=\frac{555555555}{45}=12345679$
Sum of digits of $N=37$.