Problem 7. The dragon has 40 piles of gold coins, and the number of coins in any two of them differs. After the dragon plundered a neighboring city and brought back more gold, the number of coins in each pile increased by either 2, 3, or 4 times. What is the smallest number of different piles of coins that could result?
Answer: 14.

Solution. Evaluation. Suppose that no more than 13 piles increased by the same factor. Then no more than 13 piles increased by a factor of 2, no more than 13 by a factor of 3, and no more than 13 by a factor of 4. Thus, the dragon has no more than 39 piles in total. Contradiction.

Therefore, there will be 14 piles where the number of coins increased by the same factor. Since the initial number of coins in these piles was different, the final number of coins will also be different.

Example. Suppose the dragon initially had piles with the following number of coins:

$$
\begin{aligned}
& \frac{12}{2}, \frac{12}{3}, \frac{12}{4}, \frac{120}{2}, \frac{120}{3}, \frac{120}{4}, \frac{1200}{2}, \frac{1200}{3}, \frac{1200}{4}, \ldots \\
& \ldots \frac{12 \overbrace{0 \ldots 0}^{12}}{2}, \frac{12 \overbrace{0 \ldots 0}^{12}}{3}, \frac{12 \overbrace{0 \ldots 0}^{12}}{4}, \frac{12 \overbrace{0 \ldots 0}^{13}}{2} .
\end{aligned}
$$

In each pile with a denominator of 2, he increased the number of coins by a factor of 2, in each pile with a denominator of 3 by a factor of 3, and in each pile with a denominator of 4 by a factor of 4. After this, he will have 14 different

$$
12 \text { zeros } 13 \text { zeros }
$$

values: $12,120,1200$

$\overbrace{0 \ldots 0}, 12 \overbrace{0 \ldots 0}$.