To solve the problem, we denote the number of weights selected from each stack by \(x\), \(y\), and \(z\), where:
- \(x\) is the number of 3 kg weights,
- \(y\) is the number of 5 kg weights,
- \(z\) is the number of 7 kg weights.
We need to find the minimum number of weights selected such that the total mass is 130 kg. This gives us the equation:
\[ 3x + 5y + 7z = 130 \]
We also want to minimize the total number of weights, given by:
\[ N = x + y + z \]
To minimize \(N\), we can start by observing how many weights we need to select from each type while trying to reach 130 kg.
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Maximize usage of the heaviest weights (7 kg): We divide 130 by 7 to find the maximum number of 7 kg weights we could use:
\[ z = \left\lfloor \frac{130}{7} \right\rfloor = 18 \text{ (maximum)} \] \[ 18 \times 7 = 126 \quad \text{(this leaves us needing 4 kg)} \]
We can achieve this using 1 weight of 3 kg and 1 weight of 5 kg, but that exceeds the target. Instead, we can use:
\[ z = 18 \quad (126 \text{ kg from 7 kg weights}) \] \[ 3x + 5y = 130 - 126 = 4 \]
The combinations that work are:
- \(x = 0\), \(y = 0\): Not possible.
- \(x = 1\), \(y = 1\): Gives 8 kg. Not possible.
- \(x = 0\), \(y = 1\): Gives 5 kg. Not possible.
- \(x = 2\), \(y = 0\): Gives 6 kg. Not possible.
- \(x = 0\), \(y = 0\): Gives 0 kg. Not possible self-check.
Continuing this way, let’s test lower combinations of \(z\):
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Maximize 7 kg weights and decrease gradually:
\(z = 17\): \[ 17 \times 7 = 119 \quad \Rightarrow \quad 130 - 119 = 11 \] We can use two weights of 5 kg and one weight of 3 kg: \[ 3x + 5y = 11 \] Where \(x = 1\) and \(y = 2\). Hence:
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Calculate total weights: \[ x + y + z = 1 + 2 + 17 = 20. \]
Continuing a little lower to find lower \(N\):
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Next maximum \(z = 16\): \[ 16 \times 7 = 112 \quad \Rightarrow \quad 130 - 112 = 18 \]
The set can be:
- \(x = 0\) and \(y = 3\) (Three, 5kg): \[0 + 3 + 16 = 19\]
After testing these lower ratios. Vilifying finally gives us spaced mathematics but consistently nurtured combinations of heavier weights.
Through several trials: The configuration:
- \(x = 1 \quad (3 kg)\)
- \(y = 2 \quad (5 kg)\)
- \(z = 17 \quad (7 kg)\)
calculates up to the minimum weights used while deducing 130 kg effectively ensuring 20 weights effectively are the right turn to achieve the total setup.
Finally: The Minimum number of weights required gives total outcome:
\[ \text{Total number of weights used}: \boxed{20} \]