In the set $ S_n \equal{} \{1, 2,\ldots ,n\}$ a new multiplication $ a*b$ is defined with the following properties:

[b](i)[/b] $ c \equal{} a * b$ is in $ S_n$ for any $ a \in S_n, b \in S_n.$
[b](ii)[/b] If the ordinary product $ a \cdot b$ is less than or equal to $ n,$ then $ a*b \equal{} a \cdot b.$
[b](iii)[/b] The ordinary rules of multiplication hold for $ *,$ i.e.:

[b](1)[/b] $ a * b \equal{} b * a$ (commutativity)
[b](2)[/b] $ (a * b) * c \equal{} a * (b * c)$ (associativity)
[b](3)[/b] If $ a * b \equal{} a * c$ then $ b \equal{} c$ (cancellation law).

Find a suitable multiplication table for the new product for $ n \equal{} 11$ and $ n \equal{} 12.$
To solve the problem, we need to define a new multiplication operation $a * b$ on the set $S_n = \{1, 2, \ldots, n\}$ that satisfies the given properties. We will construct the multiplication tables for $n = 11$ and $n = 12$.

### For \( n = 11 \):

1. **Define the function \( f \)**:
   We define a function \( f: S_{11} \rightarrow \mathbb{Z}/(11\mathbb{Z}) \) as follows:
   \[
   \begin{aligned}
   &f(1) = 11, \\
   &f(2) = 1, \\
   &f(3) = 5, \\
   &f(4) = 2, \\
   &f(5) = 7, \\
   &f(6) = 6, \\
   &f(7) = 4, \\
   &f(8) = 3, \\
   &f(9) = 10, \\
   &f(10) = 8, \\
   &f(11) = 9.
   \end{aligned}
   \]

2. **Define the new multiplication \( a * b \)**:
   We define \( a * b = f^{-1}(f(a) + f(b)) \mod 11 \).

3. **Verify the properties**:
   - **Commutativity**: \( a * b = b * a \) because addition in \( \mathbb{Z}/(11\mathbb{Z}) \) is commutative.
   - **Associativity**: \( (a * b) * c = a * (b * c) \) because addition in \( \mathbb{Z}/(11\mathbb{Z}) \) is associative.
   - **Cancellation law**: If \( a * b = a * c \), then \( f(a) + f(b) \equiv f(a) + f(c) \mod 11 \), which implies \( f(b) \equiv f(c) \mod 11 \), hence \( b = c \).

4. **Construct the multiplication table**:
   Using the function \( f \) and its inverse \( f^{-1} \), we construct the multiplication table for \( n = 11 \):

\[
\begin{array}{c|ccccccccccc}
* & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\
\hline
1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\
2 & 2 & 4 & 6 & 8 & 10 & 5 & 3 & 7 & 1 & 11 & 9 \\
3 & 3 & 6 & 9 & 5 & 2 & 1 & 11 & 10 & 7 & 4 & 8 \\
4 & 4 & 8 & 5 & 7 & 11 & 10 & 6 & 3 & 2 & 9 & 1 \\
5 & 5 & 10 & 2 & 11 & 8 & 4 & 1 & 9 & 6 & 7 & 3 \\
6 & 6 & 5 & 1 & 10 & 4 & 2 & 9 & 11 & 3 & 8 & 7 \\
7 & 7 & 3 & 11 & 6 & 1 & 9 & 10 & 5 & 8 & 2 & 4 \\
8 & 8 & 7 & 10 & 3 & 9 & 11 & 5 & 6 & 4 & 1 & 2 \\
9 & 9 & 1 & 7 & 2 & 6 & 3 & 8 & 4 & 11 & 5 & 10 \\
10 & 10 & 11 & 4 & 9 & 7 & 8 & 2 & 1 & 5 & 3 & 6 \\
11 & 11 & 9 & 8 & 1 & 3 & 7 & 4 & 2 & 10 & 6 & 5
\end{array}
\]

### For \( n = 12 \):

1. **Define the function \( f \)**:
   We define a function \( f: S_{12} \rightarrow \mathbb{Z}/(12\mathbb{Z}) \) as follows:
   \[
   \begin{aligned}
   &f(1) = 12, \\
   &f(2) = 1, \\
   &f(3) = 4, \\
   &f(4) = 2, \\
   &f(5) = 9, \\
   &f(6) = 5, \\
   &f(7) = 7, \\
   &f(8) = 3, \\
   &f(9) = 8, \\
   &f(10) = 10, \\
   &f(11) = 11, \\
   &f(12) = 6.
   \end{aligned}
   \]

2. **Define the new multiplication \( a * b \)**:
   We define \( a * b = f^{-1}(f(a) + f(b)) \mod 12 \).

3. **Verify the properties**:
   - **Commutativity**: \( a * b = b * a \) because addition in \( \mathbb{Z}/(12\mathbb{Z}) \) is commutative.
   - **Associativity**: \( (a * b) * c = a * (b * c) \) because addition in \( \mathbb{Z}/(12\mathbb{Z}) \) is associative.
   - **Cancellation law**: If \( a * b = a * c \), then \( f(a) + f(b) \equiv f(a) + f(c) \mod 12 \), which implies \( f(b) \equiv f(c) \mod 12 \), hence \( b = c \).

4. **Construct the multiplication table**:
   Using the function \( f \) and its inverse \( f^{-1} \), we construct the multiplication table for \( n = 12 \):

\[
\begin{array}{c|cccccccccccc}
* & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
2 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 & 22 & 24 \\
3 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 & 30 & 33 & 36 \\
4 & 4 & 8 & 12 & 16 & 20 & 24 & 28 & 32 & 36 & 40 & 44 & 48 \\
5 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 & 55 & 60 \\
6 & 6 & 12 & 18 & 24 & 30 & 36 & 42 & 48 & 54 & 60 & 66 & 72 \\
7 & 7 & 14 & 21 & 28 & 35 & 42 & 49 & 56 & 63 & 70 & 77 & 84 \\
8 & 8 & 16 & 24 & 32 & 40 & 48 & 56 & 64 & 72 & 80 & 88 & 96 \\
9 & 9 & 18 & 27 & 36 & 45 & 54 & 63 & 72 & 81 & 90 & 99 & 108 \\
10 & 10 & 20 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & 100 & 110 & 120 \\
11 & 11 & 22 & 33 & 44 & 55 & 66 & 77 & 88 & 99 & 110 & 121 & 132 \\
12 & 12 & 24 & 36 & 48 & 60 & 72 & 84 & 96 & 108 & 120 & 132 & 144
\end{array}
\]

The final answer is the multiplication tables for \( \boxed{ n = 11 } \) and \( n = 12 \) as constructed above.