Given U = \1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\. Let 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)] and solve 32y - 4 x 3^

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

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

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

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

• Set $C = \{x: x \text{ is even and not divisible by } 5\}$ The even numbers in U are 2, 4, 6, 8, 10, 12. From these, the numbers not divisible by 5 are 2, 4, 6, 8, 12 (10 is divisible by 5).

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

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

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

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 $\boxed{4}$.

Substitute $u$ into the equation to form a quadratic equation.

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

Factor the quadratic equation: We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3.

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

Set each factor to zero to find the possible values for $u$:

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

Case 1: $u = 1$

$$3^y = 1$$

Since any non-zero number raised to the power of 0 is 1, we have:

$$3^y = 3^0$$ $$y = 0$$

Case 2: $u = 3$

$$3^y = 3$$

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

$$3^y = 3^1$$ $$y = 1$$

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