A palace has gold, silver, and bronze coins in quantities of 18000, 9600, and 3600 respectively. All coins must be packed into rooms so that each room contains the same total number of coins, and each room holds coins of only one type. What is the minimum number of rooms required under these conditions?

Introduction / Context:
This problem is a practical application of the highest common factor (HCF) in the context of distributing objects evenly. The palace has three types of coins in large quantities, and the goal is to minimize the number of rooms while keeping the number of coins per room the same and using only one type of coin in each room. Such questions blend HCF with logical interpretation of distribution conditions.

Given Data / Assumptions:

• Gold coins = 18000
• Silver coins = 9600
• Bronze coins = 3600
• Each room contains coins of only one type.
• Every room has the same total number of coins.
• We want the minimum number of rooms, which means the maximum possible coins per room.

Concept / Approach:
Since each room must have the same number of coins and all three totals must be split into equal sized groups, the total number of coins per room must be a common divisor of 18000, 9600, and 3600. To minimize the number of rooms, we maximize the coins per room, so we choose the highest common factor of the three quantities. Then we compute how many rooms are needed for each type and add them together.

Step-by-Step Solution:
Find HCF of 18000, 9600, and 3600. 18000 = 2^4 * 3^2 * 5^3 9600 = 2^7 * 3^1 * 5^2 3600 = 2^4 * 3^2 * 5^2 Minimum powers for common primes: 2^4, 3^1, 5^2. HCF = 2^4 * 3 * 5^2 = 16 * 3 * 25 = 1200 coins per room. Gold rooms = 18000 / 1200 = 15. Silver rooms = 9600 / 1200 = 8. Bronze rooms = 3600 / 1200 = 3. Total rooms required = 15 + 8 + 3 = 26.

Verification / Alternative check:
Each room has 1200 coins. This fits exactly into each type total and respects the rule that a room cannot mix coin types. Any larger number of coins per room would not divide all three totals simultaneously, so would violate the equal room size condition. Hence 26 rooms is minimal.

Why Other Options Are Wrong:
24, 18, 12, and 30 correspond either to smaller common divisors, different interpretations, or arrangements with unequal numbers of coins per room. They do not satisfy all the given constraints while also minimizing the number of rooms.

Common Pitfalls:
Some learners misinterpret the condition and try to make the number of rooms the HCF directly, or they allow each type of coin to have a different coin count per room, which breaks the given requirement. The key is that every room must have the same number of coins and that no room mixes coin types, so the group size must be the HCF of all three totals.

Final Answer:
26