April wrote: For a positive integer $ n$, let $ S(n)$ denote the sum of its digits. Find the largest possible value of the expression $ \frac {S(n)}{S(16n)}$. First notice that $ S(10^kn) = S(n)$ so we see: $ \frac {S(n)}{S(16n)} = \frac {S(10000n)}{S(16n)}$ So we can assume $ 625 \mid n$. Let $ n = 625k$. Then $ \frac{S(n)}{S(16n)} = \frac{S(625 \cdot k)}{S(10000k)} = \frac{S(625 \cdot k)}{S(k)}$. For $ k = 1$ we see that $ \frac{S(625)}{S(1)} = 13$. And we just have to prove the obvious $ S(a) \cdot S(b) \ge S(ab)$. Then $ \frac{S(625 \cdot k)}{S(k)} \le S(625) = 13$. So the max is $ \boxed{13}$

$S(a)S(b) \geq S(ab) $ doesnt seem so obvious. Anyone can give a proof?

erbed wrote: $S(a)S(b) \geq S(ab) $ doesnt seem so obvious. Anyone can give a proof? Let $a$ resp. $b$ have decimal representation $a_0 \ldots a_n$ resp. $b_0 \ldots b_m$, i.e. \[ a = \sum_{i = 0}^n 10^i a_i, b = \sum_{i = 0}^m 10^i b_i, a_i,b_i \in \{0,1,\ldots,9\} \] Then $S(a)S(b) = \sum_{i = 0}^n \sum_{j = 0}^m a_ib_j$, on the other hand: \[ ab = \sum_{i = 0}^n \sum_{j = 0}^m 10^{i+j} a_i b_j \] So: \[ S(a)S(b) = \sum_{i = 0}^n \sum_{j = 0}^m S(10^{i+j} a_i b_j) \ge S \left ( \sum_{i = 0}^n \sum_{j = 0}^m 10^{i+j} a_i b_j \right ) = S(ab) \] At the last step I use $S(x) + S(y) \ge S(x+y)$ which you can easily prove yourself.

We claim that the answer is 13, achieved by 625. We wrote $16n$ as a sum of $S(16n)$ powers of 10. When this is divided by 16, it becomes a sum of a bunch of "625"s that occupy consecutive place values (some of them possibly going into decimals). Therefore, the new sum of digits can be at most 13 times the original, since each term $10^a$ becomes $6\cdot 10^{a-2}+2\cdot 10^{a-3}+5\cdot 10^{a-4}$ which contributes 13, but there could possibly be "carries" that further decrease the sum of digits, so we are done