Let's solve this problem step by step.

We have three boxes with weights $x$, $y$, and $z$. The combined weights of each pair of boxes are:

$x + y = 65 \, \text{kg}, \quad y + z = 67 \, \text{kg}, \quad z + x = 68 \, \text{kg}$

Step 1: Add all the equations together

$(x + y) + (y + z) + (z + x) = 65 + 67 + 68$ Simplifying the left side: $2(x + y + z) = 200$ Dividing by 2: $x + y + z = 100$

Step 2: Solve for individual weights

Now, subtract each of the original equations from this sum:

• To find $z$, subtract $x + y = 65$ from $x + y + z = 100$: $z = 100 - 65 = 35 \, \text{kg}$

• To find $x$, subtract $y + z = 67$ from $x + y + z = 100$: $x = 100 - 67 = 33 \, \text{kg}$

• To find $y$, subtract $z + x = 68$ from $x + y + z = 100$: $y = 100 - 68 = 32 \, \text{kg}$

Step 3: Identify the lightest box

The three weights are $x = 33 \, \text{kg}$, $y = 32 \, \text{kg}$, and $z = 35 \, \text{kg}$. The lightest box weighs 32 kg.

The answer is $\boxed{32}$.

Would you like further details on this solution? Here are 5 related questions you might find interesting:

1. How would the solution change if the combined weights of the pairs were different?
2. Can you verify the solution by adding the pairwise sums again?
3. What would happen if there were 4 boxes instead of 3?
4. How can we solve this problem using matrix equations?
5. Can we generalize this method for solving problems with more than 3 objects?

Tip: Always double-check your arithmetic to avoid simple errors, especially when summing or subtracting values.