3 rice sacks contain 120 kg, 125 kg, 100 kg rice. what is the max. amount (in weight) of rice that can be filled in small sacks so that no rice left

To determine the maximum amount of rice that can be filled into small sacks without leaving any leftover rice, we need to find the **greatest common divisor (GCD)** of the three given amounts: 120 kg, 125 kg, and 100 kg.

### Steps:
1. **Find the GCD of 120 and 100**:
   - Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
   - Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100
   - The GCD of 120 and 100 is **20**.

2. **Now find the GCD of 20 and 125**:
   - Factors of 125: 1, 5, 25, 125
   - The GCD of 20 and 125 is **5**.

Thus, the maximum weight of rice that can be packed into small sacks, so that no rice is left, is **5 kg**.

### Answer:
The maximum amount of rice that can be filled in small sacks is **5 kg** per sack.

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the solution change if the sack weights were 130 kg, 135 kg, and 110 kg?
2. What is the importance of the greatest common divisor (GCD) in packing problems like this?
3. How can you quickly find the GCD of two numbers using the Euclidean algorithm?
4. What if the small sack sizes could only hold a maximum of 10 kg—how would this affect the result?
5. What other real-life situations can benefit from the GCD method?

**Tip**: The Euclidean algorithm is a fast way to compute the GCD of two numbers!

3 rice sacks contain 120 kg, 125 kg, 100 kg rice. what is the max. amount (in weight) of rice that can be filled in small sacks so that no rice left?

Discover how to solve the problem of distributing 120 kg, 125 kg, and 100 kg of rice into small sacks without leftovers by using the Greatest Common Divisor (GCD). This solution applies the Euclidean Algorithm and is ideal for students in Grades 6-8.

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