Question

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?

Answer 1

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:

\[
\text{Total number of weights used}: \boxed{20}
\]