If $S(m)=20$ and $S(33m)=120$, what is the value of $S(3m)$? Let $S(n)$ denotes the sum of the digits of a positive integer $n$ in base $10$.
 If $S(m)=20$ and $S(33m)=120$, what is the value of $S(3m)$?
I do not even have a start, so a start would be appreciated or even a solution would be okay.
 A: The point is the following : the way the question is written, it seems that $s(3m)$ is independent of the choice of $m$, if $s(m)$ and $s(33m)$ are as given. Therefore, it seems that the following has can be done : find a number $m$ satisfying the conditions, and then explicitly find $s(3m)$.
NOTE : A proof has been given in the other answer : I have preferred to actually provide a candidate for a number $m$, for which the quantity $s(3m)$ is easy to calculate.
The good news for us is that $s(33m) = s(33)s(m)$, since $20 \times 6= 120$. This immediately gives a candidate for $m$ : when me multiply a number looking like $1111$ with say $8$, we get $8888$, whose digit sum is easy to calculate.
In the same vein, consider :
$$
m = 101010101010101010101010101010101010101
$$
Then $m$ contains $20$ ones and $19$ zeros, so $s(m) = 20$. Also :
$$
33m = 3333333333333333333333333333333333333333
$$
Which has forty digits each of which is $3$, so $s(33m) = 120$. 
It is easy to see what $s(3m)$ is :
$$
303030303030303030303030303030303030303
$$
And so $s(3m) = 60$.
A: Observe that $S(a)+S(b)\ge S(a+b)$ and $S(10a)=S(a)$ for any $a,b\in\mathbb{N}$.
$S(m)=20\Rightarrow S(3m)\le 3S(m)=60$. $S(33m)=120\Rightarrow 120\le S(3m\cdot 10)+S(3m)=S(3m)+S(3m)\Rightarrow S(3m)\ge 60$.
$\therefore S(3m)=60$
