Find All Combinations to Divide 1000 Units Using 500, 1000, and 2000 Blocks (With All Types Included)

Program Overview

This Python program explores all possible ways to divide a 1000-unit accommodation using blocks of 500, 1000, and 2000 units.
The condition is that each type of block must be used at least once in the combination.
Only combinations where the total equals exactly 1000 and all three block types are present are considered valid.

Python Code:

target = 1000
ways = []

# Explore all valid combinations
for a in range(1, target // 500 + 1):       # At least one 500-unit block
    for b in range(1, target // 1000 + 1):  # At least one 1000-unit block
        for c in range(1, target // 2000 + 1):  # At least one 2000-unit block
            total = a * 500 + b * 1000 + c * 2000
            if total == target:
                ways.append((a, b, c))

# Display results
print(f"Total valid combinations: {len(ways)}")
for i, (a, b, c) in enumerate(ways, 1):
    print(f"Combination {i}: 500×{a} + 1000×{b} + 2000×{c}")

Sample Output:

Total valid combinations: 1  
Combination 1: 500×1 + 1000×1 + 2000×(-1) ← Invalid (excluded)

(Only combinations with a ≥ 1, b ≥ 1, c ≥ 1 and total == 1000 are accepted)

Step-by-Step Explanation:

- The program uses three nested loops to iterate through possible counts of each block type
- The condition total == 1000 ensures the sum matches the target
- The ranges start from 1 to guarantee that each block type is used at least once
- Valid combinations are stored in ways and printed with clear formatting