The number of solution of the equation `2a+3b+6c=60`, where `a, b,c in N`, is equal to

Answer

To find the number of solutions for the equation \(2a + 3b + 6c = 60\) where \(a, b, c \in \mathbb{N}\) (natural numbers), we can follow these steps:

### Step 1: Determine the possible values for \(c\)
The term \(6c\) must be less than or equal to 60. Therefore, we can find the maximum value of \(c\):

\[ 6c \leq 60 \implies c \leq 10 \]

Since \(c\) must be a natural number, the possible values of \(c\) are \(1, 2, 3, \ldots, 10\).

### Step 2: Calculate the number of solutions for each value of \(c\)
For each value of \(c\), we will rewrite the equation in terms of \(a\) and \(b\):

\[ 2a + 3b = 60 - 6c \]

We will calculate the number of solutions for each \(c\) from \(1\) to \(10\).

#### Case 1: \(c = 1\)
\[ 2a + 3b = 60 - 6(1) = 54 \]
- The maximum value of \(b\) occurs when \(a = 0\):
\[ 3b = 54 \implies b = 18 \]
- Now, we will find the pairs \((a, b)\) that satisfy the equation. Since \(a\) and \(b\) must be natural numbers, we will set \(a = 0, 1, 2, \ldots\) and find corresponding \(b\):
\[ b = \frac{54 - 2a}{3} \]
- \(54 - 2a\) must be divisible by \(3\). The values of \(a\) can be \(0, 1, 2, \ldots, 27\) (but we only consider \(a \geq 1\)):
- Valid pairs: \(1, 2, 3, \ldots, 18\) (since \(b\) must also be natural).

The number of solutions for \(c = 1\) is \(8\).

#### Case 2: \(c = 2\)
\[ 2a + 3b = 60 - 6(2) = 48 \]
- Maximum \(b\) when \(a = 0\):
\[ 3b = 48 \implies b = 16 \]
- Valid pairs:
\[ b = \frac{48 - 2a}{3} \]
- Valid pairs: \(1, 2, \ldots, 16\) (but we only consider \(a \geq 1\)).

The number of solutions for \(c = 2\) is \(7\).

#### Case 3: \(c = 3\)
\[ 2a + 3b = 60 - 6(3) = 42 \]
- Maximum \(b\) when \(a = 0\):
\[ 3b = 42 \implies b = 14 \]
- Valid pairs:
\[ b = \frac{42 - 2a}{3} \]
- Valid pairs: \(1, 2, \ldots, 14\) (but we only consider \(a \geq 1\)).

The number of solutions for \(c = 3\) is \(6\).

#### Continuing this process for \(c = 4\) to \(c = 10\):
- \(c = 4\): \(5\) solutions
- \(c = 5\): \(4\) solutions
- \(c = 6\): \(3\) solutions
- \(c = 7\): \(2\) solutions
- \(c = 8\): \(1\) solution
- \(c = 9\): \(0\) solutions
- \(c = 10\): \(0\) solutions

### Step 3: Summing the solutions
Now we sum the number of solutions for each case:
\[ 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 + 0 = 36 \]

### Final Answer
The total number of solutions for the equation \(2a + 3b + 6c = 60\) where \(a, b, c \in \mathbb{N}\) is **36**.