Digit sum of $3k$ given the digit sums of $k$ and $44k$

Question

Define $S(n)$ to be the digit sum of $n,n\in\mathbb{N_{\geqslant0}}$. Then it is given that:

1. $S(n)\equiv n\pmod9,\forall n\in\mathbb{N}$
2. $S(a+b)\leqslant S(a)+S(b),\forall a,b\in\mathbb{N}$
3. $S(ab)\leqslant S(a)S(b),\forall a,b\in\mathbb{N}$

Determine $S(3k)$ given that $S(k)=100$ and $S(44k)=800$, $k\in\mathbb{N_{\geqslant0}}$. Do not find $k$.

This is what I have gathered so far:

1. $S(3)=3$ and $S(k)=100$
2. $S(3k)\leqslant S(3)\times S(k)$
3. $\therefore S(3k)\leqslant 300$

From here on I'm not sure how to determine the precise value of $S(k)$. I know that a possible $k$ for $S(k)=100$ and $S(44k)=800$ is $k=\frac{10^{100}-1}{9}=11\cdots1$, where $k$ has 100 digits and all its digits are $1$. Then $S(\frac{10^{100}-1}{9})=300$. But the answer should not include the value of $k$.

Any ideas how I can change the above to become an equality from an inequality?

Answer 1

We can prove $S(3k) \geqslant 300$ (and hence equality) as follows:

Since $44k = 11 \cdot 4k$,

\begin{align*} 800 = S(44k) \leqslant 2 \cdot S(4k), \end{align*} so $S(4k) \geqslant 400$. But since

\begin{align*} S(4k) = S(3k + k) \leqslant S(3k) + S(k) = S(3k) + 100, \end{align*} we get that

\begin{align*} S(3k) \geqslant S(4k) - 100 \geqslant 300. \end{align*}