To measure exactly 31 kg of rice on a balance scale, including a 1 kg standard weight you already have, what is the minimum number of distinct weights required?

Introduction / Context:
This arithmetic reasoning puzzle is based on the idea of representing a given weight using a set of standard weights on a balance scale. The goal is to find the minimum number of distinct weights required, given that we already have a 1 kg weight, so that we can measure exactly 31 kg of rice. This relates to representing numbers in terms of powers of 2 on a balance scale in a straightforward way where all weights are placed on one side and the object is on the other side.

Given Data / Assumptions:
- We need to measure exactly 31 kg of rice.- We have a balance scale and standard weights.- One 1 kg weight is already available.- We assume we place all standard weights on one pan and the rice on the other pan, not allowing weights on both sides.- We want to minimize the total number of distinct standard weights including the existing 1 kg weight.

Concept / Approach:
To measure any integer weight up to some maximum using weights all placed on one side, a very efficient system is based on powers of 2. Any positive integer can be written as a sum of distinct powers of 2, which is the idea behind binary representation. To express 31 as a sum of powers of 2, we write 31 = 16 + 8 + 4 + 2 + 1. Thus, the required weights are 1 kg, 2 kg, 4 kg, 8 kg, and 16 kg. Since 1 kg is already available, we only need to ensure the complete set of these five distinct weights.

Step-by-Step Solution:
- Express 31 in binary or as a sum of powers of 2.- 31 = 16 + 8 + 4 + 2 + 1.- The powers of 2 involved are 1, 2, 4, 8, and 16.- Therefore, you need weights of 1 kg, 2 kg, 4 kg, 8 kg, and 16 kg.- The total number of distinct weights is 5.- With these weights on one pan, you can place the rice on the other pan and combine the weights to get 31 kg exactly.

Verification / Alternative check:
Check that no smaller set works. If you try to remove any one of the powers of 2, you can no longer construct 31 using only sums of the remaining weights. For example, without the 16 kg weight, the largest sum you can make from 1, 2, 4, and 8 kg is 15 kg, which is not enough. Similarly, you cannot reach 31 kg without all five powers of 2. Hence 5 is minimal.

Why Other Options Are Wrong:
- Option 3: With only three weights you cannot represent 31 kg as a sum of distinct weights including 1 kg.- Option 7: Seven weights are more than necessary and not minimal.- Option 9: Nine weights are far too many for such a simple target.- Option 6: Six weights would include redundant values and are not needed because five already achieve the goal.

Common Pitfalls:
A common mistake is to think in terms of repeated uses of a single weight, for example imagining weighing 1 kg thirty one times. In such puzzles, each standard weight is used at most once in a given weighing. Another error is to forget the binary idea and choose arbitrary weights that do not combine efficiently. Remember that powers of 2 are optimal for covering a range of integer values on a balance with all weights on one side.

Final Answer:
The minimum number of distinct weights needed, including the 1 kg weight, is 5.