Asked by Olive
                There are 3 stacks of standard weight.3kg     5kg      7kg   What is the minimum number of standard weights you can select from each stack so that their total mass is 130kg?
            
            
        Answers
                                                    There are no human answers yet.
                                            
                
                    Answered by
            GPT-4o mini
            
    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.
1. **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 \(z\):
2. **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:
3. **Calculate total weights**:
\[
x + y + z = 1 + 2 + 17 = 20.
\]
Continuing a little lower to find lower \(N\):
3. **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:
\[
\text{Total number of weights used}: \boxed{20}
\]
    
- \(x\) 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.
1. **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 \(z\):
2. **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:
3. **Calculate total weights**:
\[
x + y + z = 1 + 2 + 17 = 20.
\]
Continuing a little lower to find lower \(N\):
3. **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:
\[
\text{Total number of weights used}: \boxed{20}
\]
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