Step-by-step Solution:

Let's define the five different boxes with weights \( a, b, c, d, e \) such that: \[ a < b < c < d < e \] Each pair of boxes was weighed, resulting in the following weights: \[ 110, 112, 113, 114, 115, 116, 117, 118, 120, 121 \] Step 1: Finding the Sum of All Weights
Each weight corresponds to a sum of two different box weights. Since there are five boxes, we have: \[ \text{Total sum of all pairwise weights} = (a+b) + (a+c) + (a+d) + (a+e) + (b+c) + (b+d) + (b+e) + (c+d) + (c+e) + (d+e) \] Adding all the given weights: \[ 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1156 \] This sum is equal to: \[ 4(a + b + c + d + e) \] Thus, \[ a + b + c + d + e = \frac{1156}{4} = 289 \] Step 2: Identifying the Smallest and Largest Weights
The smallest sum must be \( a + b = 110 \) and the largest sum must be \( d + e = 121 \).
Since we know the total sum: \[ a + b + c + d + e = 289 \] We substitute \( a + b = 110 \) and \( d + e = 121 \): \[ 110 + c + 121 = 289 \] \[ c = 289 - 231 = 58 \] Step 3: Identifying the Heaviest Box
The heaviest box is \( e \), which is obtained from: \[ d + e = 121 \] From other pairwise sums, we find \( d \) and \( e \):
\( c + e = 120 \Rightarrow 58 + e = 120 \Rightarrow e = 62 \)
\( c + d = 118 \Rightarrow 58 + d = 118 \Rightarrow d = 60 \)
Thus, the heaviest box weighs: \( \mathbf{62 \text{ kg}} \)