Baltic Way 2025 · Problem 16

We use three lemmas.

Lemma 1. For positive integers $p,q$,

$$S(p+q)\le S(p)+S(q).$$

Proof. In fact,

$$S(p+q)=S(p)+S(q)-9\cdot(\text{number of carries when adding }p\text{ and }q),$$

which follows directly from decimal addition.

Lemma 2. For positive integers $p,q$,

$$S(pq)\le S(p)S(q).$$

Proof. Write

$$p=p_m10^m+p_{m-1}10^{m-1}+\cdots+p_0, \qquad 0\le p_i\le9.$$

Using Lemma 1 repeatedly,

$$\begin{aligned} S(qp) &\le S(qp_m10^m)+S(qp_{m-1}10^{m-1})+\cdots+S(qp_0)\\ &=S(qp_m)+S(qp_{m-1})+\cdots+S(qp_0)\\ &\le p_mS(q)+p_{m-1}S(q)+\cdots+p_0S(q)\\ &=S(p)S(q). \end{aligned}$$

Lemma 3. For every positive integer $m$, every positive multiple of $10^m-1$ has digit sum at least $9m$.

Proof. We induct on the multiple. Certainly $S(10^m-1)=9m$. Let $p$ be a larger multiple and write

$$p=10^mp_1+p_0, \qquad 0\le p_0<10^m.$$

Since $10^m\equiv1\pmod{10^m-1}$, the integer $p_1+p_0$ is also divisible by $10^m-1$, and it is smaller than $p$. Hence

$$S(p)=S(p_1)+S(p_0)\ge S(p_1+p_0)\ge9m.$$

Returning to the problem, $ab^2c^4$ is a multiple of $abc=10^n-1$, so Lemma 3 gives

$$S(ab^2c^4)\ge9n.$$

By AM-GM and Lemma 2,

$$\begin{aligned} S(a)+S(b^2)+S(c^4) &\ge3\sqrt[3]{S(a)S(b^2)S(c^4)}\\ &\ge3\sqrt[3]{S(ab^2c^4)}\\ &\ge3\sqrt[3]{9n} =\sqrt[3]{243n}. \end{aligned}$$