Solution

The removal of one of the pots should allow Colorado Jones to separate the remaining ones into two groups, where in one of these groups there is double the number of coins of the other group. If the number of coins in the smaller group is $x$, the sum of the coins in the two groups is $x+2x=3x$. Therefore, we already know that removing the correct pot makes the remaining number of coins a multiple of 3. The sum of the number of coins in the five pots is $81+71+41+37+35=265$, which leaves a remainder of 1 when divided by 3. Consequently, to make this quantity a multiple of 3 by removing a pot, the pot must leave a remainder of 1 when divided by 3. Among the numbers given, only 37 has this property. Furthermore, see that by removing it, the remaining pots can be divided into two groups: one with a sum of $81+71=2 \cdot 76$ and the other with a sum of $35+41=76$. Thus, Colorado Jones should remove the pot with 37 coins.
Colorado Jones must solve a great riddle to survive. He must remove only one of the five pots in front of him, as indicated in the figure below, to open the door of the secret chamber. He knows that each pot contains only one type of coin, gold or silver, and that each number written on them represents the number of coins inside. Additionally, the only correct pot to be removed is the one that, when removed, leaves the remaining pots with the number of silver coins being double the number of gold coins. Which pot should be removed?

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-07.jpg?height=125&width=708&top_left_y=1231&top_left_x=777)

#