We have 1386 coins and a double-bottomed scale. One of the coins is fake and its weight is less than the others. In the worst circumstances, at least how many times can you find the coin by using the scale?

To solve this problem, we need to determine the minimum number of weighings required to find the single fake coin among 1386 coins using a double-bottomed (balance) scale, where the fake coin is lighter.

### Key Idea:
In each weighing with a balance scale, we can compare two groups of coins. There are three possible outcomes:
1. The left side is heavier.
2. The right side is heavier.
3. Both sides weigh the same.

This three-outcome process implies that, with each weighing, we can reduce the number of possible coins by a factor of 3. So, after each weighing, we eliminate two-thirds of the remaining possibilities. Therefore, the question boils down to determining how many weighings are needed to narrow the search to 1 coin from 1386.

### Approach:
Let \( n \) be the number of weighings required. After \( n \) weighings, we need to have narrowed the possibilities down to 1 coin. Since each weighing gives 3 outcomes, the total number of possibilities after \( n \) weighings is \( 3^n \).

Thus, we need to find the smallest \( n \) such that:
\[
3^n \geq 1386
\]

### Calculating \( n \):

We start by calculating powers of 3:
\[
3^1 = 3
\]
\[
3^2 = 9
\]
\[
3^3 = 27
\]
\[
3^4 = 81
\]
\[
3^5 = 243
\]
\[
3^6 = 729
\]
\[
3^7 = 2187
\]

Thus, \( 3^7 = 2187 \) is the smallest power of 3 that exceeds 1386. Therefore, \( n = 7 \).

### Conclusion:
In the worst-case scenario, you will need at least **7 weighings** to find the fake coin.

Would you like any further details or clarifications?

Here are 5 related questions to expand on the topic:
1. How does the process change if the fake coin can be either lighter or heavier?
2. How would the strategy differ if we had more than one fake coin?
3. What if we were using a digital scale instead of a balance scale—how would that affect the number of weighings?
4. How would the process differ if we had only 2 possible outcomes per weighing (instead of 3)?
5. What if the number of coins was significantly larger, say 10,000? How many weighings would you need then?

**Tip:** Always try to group coins in approximately equal parts when using a balance scale for optimal efficiency in narrowing down the possibilities.

Discover the minimum number of weighings needed to find a single fake coin among 1386 using a balance scale. This problem uses concepts from combinatorics and logarithmic growth, suitable for students in Grades 10-12.

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