To find the number of solutions for the equation \(2a + 3b + 6c = 60\) where \(a, b, c \in \mathbb{N}\) (natural numbers), we will analyze the equation step by step.
### Step 1: Determine the possible values of \(c\)
The term \(6c\) must be less than or equal to \(60\). Thus, we can find the maximum value of \(c\):
\[
6c \leq 60 \implies c \leq 10
\]
So, \(c\) can take values from \(1\) to \(10\).
### Step 2: Analyze the equation for each value of \(c\)
We will substitute values of \(c\) from \(1\) to \(10\) and find the corresponding values of \(a\) and \(b\).
#### For \(c = 1\):
\[
2a + 3b + 6(1) = 60 \implies 2a + 3b = 54
\]
- When \(a = 0\), \(3b = 54 \implies b = 18\)
- Decreasing \(b\) by \(2\) (to account for \(a\)) gives values: \(b = 16, 14, 12, 10, 8, 6, 4, 2\) (total of 8 values).
So, for \(c = 1\), there are \(8\) solutions.
#### For \(c = 2\):
\[
2a + 3b + 6(2) = 60 \implies 2a + 3b = 48
\]
- When \(a = 0\), \(3b = 48 \implies b = 16\)
- Decreasing \(b\) gives values: \(b = 14, 12, 10, 8, 6, 4, 2\) (total of 7 values).
So, for \(c = 2\), there are \(7\) solutions.
#### For \(c = 3\):
\[
2a + 3b + 6(3) = 60 \implies 2a + 3b = 42
\]
- When \(a = 0\), \(3b = 42 \implies b = 14\)
- Decreasing \(b\) gives values: \(b = 12, 10, 8, 6, 4, 2\) (total of 6 values).
So, for \(c = 3\), there are \(6\) solutions.
#### For \(c = 4\):
\[
2a + 3b + 6(4) = 60 \implies 2a + 3b = 36
\]
- When \(a = 0\), \(3b = 36 \implies b = 12\)
- Decreasing \(b\) gives values: \(b = 10, 8, 6, 4, 2\) (total of 5 values).
So, for \(c = 4\), there are \(5\) solutions.
#### For \(c = 5\):
\[
2a + 3b + 6(5) = 60 \implies 2a + 3b = 30
\]
- When \(a = 0\), \(3b = 30 \implies b = 10\)
- Decreasing \(b\) gives values: \(b = 8, 6, 4, 2\) (total of 4 values).
So, for \(c = 5\), there are \(4\) solutions.
#### For \(c = 6\):
\[
2a + 3b + 6(6) = 60 \implies 2a + 3b = 24
\]
- When \(a = 0\), \(3b = 24 \implies b = 8\)
- Decreasing \(b\) gives values: \(b = 6, 4, 2\) (total of 3 values).
So, for \(c = 6\), there are \(3\) solutions.
#### For \(c = 7\):
\[
2a + 3b + 6(7) = 60 \implies 2a + 3b = 18
\]
- When \(a = 0\), \(3b = 18 \implies b = 6\)
- Decreasing \(b\) gives values: \(b = 4, 2\) (total of 2 values).
So, for \(c = 7\), there are \(2\) solutions.
#### For \(c = 8\):
\[
2a + 3b + 6(8) = 60 \implies 2a + 3b = 12
\]
- When \(a = 0\), \(3b = 12 \implies b = 4\)
- Decreasing \(b\) gives values: \(b = 2\) (total of 1 value).
So, for \(c = 8\), there is \(1\) solution.
#### For \(c = 9\) and \(c = 10\):
For \(c = 9\):
\[
2a + 3b + 6(9) = 60 \implies 2a + 3b = 6
\]
- When \(a = 0\), \(3b = 6 \implies b = 2\)
- Decreasing \(b\) gives no valid solutions.
So, for \(c = 9\), there are \(0\) solutions.
For \(c = 10\):
\[
2a + 3b + 6(10) = 60 \implies 2a + 3b = 0
\]
- No valid solutions since \(a\) and \(b\) must be natural numbers.
### Step 3: Summing the solutions
Now, we can sum the number of solutions for each value of \(c\):
\[
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 + 0 = 36
\]
### Conclusion
The total number of solutions for the equation \(2a + 3b + 6c = 60\) where \(a, b, c \in \mathbb{N}\) is \(36\).
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