An intelligent trader travels from 1 place to another carrying 3 sacks having 30 coconuts each. No sack can hold more than 30 coconuts. On the way he passes through 30 checkpoints and on each checkpoint he has to give 1 coconut for each sack he is carrying. How many coconuts are left in the end?

This problem involves optimizing the number of coconuts the trader can have at the end after passing through 30 checkpoints, where he has to give away coconuts depending on the number of sacks he is carrying.

### Breakdown of the Problem:

- The trader starts with 3 sacks, each holding 30 coconuts. Thus, he has \(3 \times 30 = 90\) coconuts initially.
- There are 30 checkpoints.
- At each checkpoint, for every sack he is carrying, he must give up 1 coconut per sack.
- If he carries fewer sacks, he gives up fewer coconuts.

### Logical Solution:

1. **Checkpoints where the trader carries 3 sacks**:
   - If the trader carries 3 sacks, he loses 3 coconuts per checkpoint.
   - But if he carries 2 sacks, he only loses 2 coconuts per checkpoint, and if he carries 1 sack, he loses 1 coconut per checkpoint.
   
2. **Phase 1: Carrying 3 sacks**:
   - The trader will not want to carry 3 sacks all the way through because it will cost him 3 coconuts per checkpoint.
   - Therefore, he should drop off coconuts along the way, so at some point he carries fewer sacks.
   - The first strategy is to go through a certain number of checkpoints while carrying 3 sacks and then drop off coconuts.

3. **Optimal number of checkpoints for carrying 3 sacks**:
   - After 10 checkpoints, the trader would lose \(3 \times 10 = 30\) coconuts.
   - After 10 checkpoints, the trader should consolidate his coconuts into 2 sacks to minimize the number of coconuts he gives at each checkpoint.

4. **Phase 2: Carrying 2 sacks**:
   - Now the trader carries 2 sacks. At each checkpoint, he loses 2 coconuts.
   - He can pass through another 10 checkpoints, losing \(2 \times 10 = 20\) coconuts.
   - After these 10 checkpoints, he can again consolidate the remaining coconuts into 1 sack.

5. **Phase 3: Carrying 1 sack**:
   - Now the trader has only 1 sack of coconuts.
   - He can pass through the remaining 10 checkpoints, losing \(1 \times 10 = 10\) coconuts.

### Calculation:

- After the first 10 checkpoints (while carrying 3 sacks), he has \(90 - 30 = 60\) coconuts.
- After the next 10 checkpoints (while carrying 2 sacks), he has \(60 - 20 = 40\) coconuts.
- After the last 10 checkpoints (while carrying 1 sack), he has \(40 - 10 = 30\) coconuts.

### Conclusion:
The trader is left with **30 coconuts** at the end of his journey.

Would you like a deeper breakdown, or have any questions? Here are 5 related questions:

1. What if the number of checkpoints was different? How would that affect the final result?
2. What if the sacks could carry more than 30 coconuts? How would this change the strategy?
3. How many checkpoints would the trader pass through if he could consolidate coconuts earlier?
4. Could the trader drop coconuts temporarily at some checkpoints and retrieve them later?
5. What is the general formula for the number of coconuts left after passing through any number of checkpoints?

Tip: When solving problems involving optimization, always look for ways to minimize the loss per step, especially when resources are involved.

An intelligent trader travels from 1 place to another carrying 3 sacks having 30 coconuts each. No sack can hold more than 30 coconuts. On the way, he passes through 30 checkpoints, and at each checkpoint, he has to give 1 coconut for each sack he is carrying. How many coconuts are left in the end?

Explore how an intelligent trader optimizes his journey with 3 sacks of coconuts through 30 checkpoints. Learn the strategy to minimize coconut loss by reducing the number of sacks carried at each stage. Suitable for high school students, this problem covers optimization, arithmetic sequences, and algebraic reasoning.

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