Out of 6561 identical looking balls, exactly one ball is heavier than all the others. Using a balance scale that can compare the weights of two groups of balls, what is the minimum number of weighings needed to be certain of finding the heavy ball?

Introduction / Context:
This is a classic puzzle about finding one heavier ball among many identical looking balls using a balance scale. It tests your ability to use divide and conquer strategies and to think in terms of base representations, specifically base 3. Such problems often appear in logical reasoning and competitive exams, especially in sections involving puzzles and optimization.

Given Data / Assumptions:
- There are 6561 balls in total.- All balls look identical, but exactly one ball is heavier than the others.- A balance scale is available which compares the total weight on its left pan to the total weight on its right pan.- Each weighing tells us whether the left side is heavier, the right side is heavier, or both sides are equal.- We want the minimum number of weighings that guarantees we can identify the heavy ball.

Concept / Approach:
The optimal strategy uses a ternary or base three approach. In each weighing, you can split the remaining candidate balls into three groups as equal in size as possible: left pan, right pan, and one group set aside. Depending on the outcome of the weighing, you can discard two thirds of the balls and keep one third for further investigation. This is because there are three possible outcomes for each weighing: left heavier, right heavier, or balanced. The idea is similar to a ternary search where the search space reduces by a factor of three each time. The number of weighings required is the smallest integer k such that 3^k is at least the total number of balls.

Step-by-Step Solution:
- Determine how many balls can be resolved with k weighings using the base three idea.- With 1 weighing, you can distinguish at most 3 possibilities (3^1).- With 2 weighings, at most 3^2 = 9 balls can be resolved.- With 3 weighings, at most 3^3 = 27 balls can be handled.- With 4 weighings, 3^4 = 81 balls.- With 5 weighings, 3^5 = 243 balls.- With 6 weighings, 3^6 = 729 balls.- With 7 weighings, 3^7 = 2187 balls.- With 8 weighings, 3^8 = 6561 balls.- Our problem has exactly 6561 balls, which equals 3^8.- Therefore, 8 weighings are sufficient to guarantee finding the heavier ball.

Verification / Alternative check:
You can imagine the process: initially divide the 6561 balls into three groups of 2187 each. Weigh two of these groups. If one side is heavier, the heavy ball is among those 2187 balls. If the scale balances, the heavy ball is among the remaining 2187 not weighed. This process can be repeated, always dividing the current candidate group into three equal parts. After each weighing, the candidate set shrinks by a factor of three. After 8 such steps, you will be left with exactly one ball that must be heavier, confirming that 8 weighings are enough.

Why Other Options Are Wrong:
- 2414 and 204: These values are far larger than the optimal and do not represent a minimal strategy.- 87: This is also much larger than needed and indicates using a naive process such as comparing very small groups repeatedly.- 10: While 10 weighings are more than enough, they are not minimal because 8 already suffice.

Common Pitfalls:
Many learners attempt a binary split strategy that divides the balls into two groups each time, which is less efficient with a three outcome scale. Another common error is to underestimate the power of exponential growth and assume that many more weighings are needed. Remember to match the number of outcomes per weighing with the base of the representation used for counting possibilities. A balance scale with three possible outcomes naturally suggests a base three strategy.

Final Answer:
The minimum number of weighings required to be certain of finding the heavy ball is 8.