Given the universal set U = \1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\ and sets A = : x is divisible by 3 B = : x is a factor of 18 and C = : x is even and not divisible by 5 find n[A (B C

$A = \{x : x \text{ is divisible by } 3\}$
The numbers in $U$ divisible by 3 are 3, 6, 9, 12.

$$A = \{3, 6, 9, 12\}$$

$B = \{x : x \text{ is a factor of } 18\}$
The factors of 18 are 1, 2, 3, 6, 9, 18. From $U$, these are 1, 2, 3, 6, 9.

$$B = \{1, 2, 3, 6, 9\}$$

$C = \{x : x \text{ is even and not divisible by } 5\}$
Even numbers in $U$: $\{2, 4, 6, 8, 10, 12\}$.
Numbers in $U$ divisible by 5: $\{5, 10\}$.
Numbers that are even and not divisible by 5: $\{2, 4, 6, 8, 12\}$.

$$C = \{2, 4, 6, 8, 12\}$$

$$B \cup C = \{1, 2, 3, 6, 9\} \cup \{2, 4, 6, 8, 12\}$$ $$B \cup C = \{1, 2, 3, 4, 6, 8, 9, 12\}$$

$$A \cap (B \cup C) = \{3, 6, 9, 12\} \cap \{1, 2, 3, 4, 6, 8, 9, 12\}$$

The common elements are 3, 6, 9, 12.

$$A \cap (B \cup C) = \{3, 6, 9, 12\}$$

The set $A \cap (B \cup C)$ has 4 elements.

$$n[A \cap (B \cup C)] = 4$$

The value of $n[A \cap (B \cup C)]$ is $\boxed{4}$.

The given equation is $3^{2y} - 4 \times 3^y + 3 = 0$.
We can write $3^{2y}$ as $(3^y)^2$.

$$(3^y)^2 - 4 \times 3^y + 3 = 0$$

Let $u = 3^y$. Substitute $u$ into the equation:

$$u^2 - 4u + 3 = 0$$

Factor the quadratic equation:

$$(u - 1)(u - 3) = 0$$

This gives two possible values for $u$:

$$u - 1 = 0 \implies u = 1$$ $$u - 3 = 0 \implies u = 3$$

Case 1: $u = 1$

$$3^y = 1$$

Since $3^0 = 1$, we have:

$$y = 0$$

Case 2: $u = 3$

$$3^y = 3$$

Since $3^1 = 3$, we have:

$$y = 1$$

The solutions for $y$ are $\boxed{0 \text{ and } 1}$.