Basic of Set theory Questions in English

(A) The set $\emptyset $ is the empty set,which contains no elements.
By definition,the empty set is a subset of every set.
Therefore,the only subset of $\emptyset $ is $\emptyset $ itself.

(B) We know that if $A$ is a set with $m$ elements,i.e.,$n(A) = m$,then the number of elements in the power set is given by $n[P(A)] = 2^{m}$.
If $A = \varnothing$,then the number of elements in $A$ is $n(A) = 0$.
Therefore,$n[P(A)] = 2^{0} = 1$.
Hence,$P(A)$ has $1$ element.

(C) The given set is $\{ x : x \in R, -4 < x \le 6 \}$.
Since the inequality $-4 < x$ is strict,the interval is open at $-4$.
Since the inequality $x \le 6$ includes equality,the interval is closed at $6$.
Therefore,the interval is $(-4, 6]$.

(C) The given set is $\{x : x \in \mathbb{R}, 0 \le x < 7\}$.
Since $0$ is included (due to $\le$) and $7$ is excluded (due to $ < $),the interval is closed at $0$ and open at $7$.
Therefore,the interval is $[0, 7)$.

(A) The set $\{ x:x \in R, 3 \le x \le 4 \}$ represents all real numbers $x$ such that $x$ is greater than or equal to $3$ and less than or equal to $4$.
Since both endpoints $3$ and $4$ are included in the set,we use closed brackets.
Therefore,the interval notation is $[3, 4]$.

(B) The interval $(-3, 0)$ represents the set of all real numbers strictly between $-3$ and $0$.
In set-builder form,this is written as:
$(-3, 0) = \{x : x \in R, -3 < x < 0\}$

(C) The interval $[6, 12]$ is a closed interval,which includes all real numbers $x$ such that $6 \le x \le 12$.
In set-builder form,this is represented as:
${x : x \in R, 6 \le x \le 12}$
Therefore,the correct option is $C$.

(C) The interval $(6, 12]$ represents all real numbers $x$ such that $6 < x \le 12$.
In set-builder form,this is written as:
$\{x : x \in R, 6 < x \le 12\}$

(N/A) The interval $\left[ -23, 5 \right)$ represents all real numbers $x$ such that $-23 \le x < 5$.
In set-builder form,this is written as: $\{ x : x \in \mathbb{R}, -23 \le x < 5 \}$

(N/A) universal set is a set that contains all the elements under consideration. For the set of right triangles,the universal set can be the set of all triangles or the set of all polygons.

(N/A) For the set of isosceles triangles,the universal set can be the set of all triangles in a plane or the set of all polygons in a plane or the set of all two-dimensional figures in a plane.

(D) universal set for a collection of sets must contain all elements present in every set of that collection.
For the given sets $A = \{1, 3, 5\}$,$B = \{2, 4, 6\}$,and $C = \{0, 2, 4, 6, 8\}$,the universal set $U$ must satisfy $A \subseteq U$,$B \subseteq U$,and $C \subseteq U$.
Checking the set $X = \{0, 1, 2, 3, 4, 5, 6\}$:
$A \subset X$ is true.
$B \subset X$ is true.
However,$C \not\subset X$ because the element $8 \in C$ but $8 \notin X$.
Therefore,the set $\{0, 1, 2, 3, 4, 5, 6\}$ is not a universal set for $A$,$B$,and $C$.

(C) universal set $U$ for a collection of sets must contain all elements present in those sets.
For the given sets $A = \{1, 3, 5\}$,$B = \{2, 4, 6\}$,and $C = \{0, 2, 4, 6, 8\}$,the universal set must contain all elements from the union $A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$.
Option $A$ is $\varnothing$,which does not contain any elements.
Option $B$ is missing $8$.
Option $D$ is missing $0$.
Option $C$ contains all elements of $A$,$B$,and $C$ (i.e.,$0, 1, 2, 3, 4, 5, 6, 8 \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$).
Therefore,the correct universal set is $C$.

(B) universal set $U$ must contain all elements of the given sets $A$,$B$,and $C$.
Given $A = \{1, 3, 5\}$,$B = \{2, 4, 6\}$,and $C = \{0, 2, 4, 6, 8\}$.
The union of these sets is $A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$.
Since $\{0, 1, 2, 3, 4, 5, 6, 8\} \subset \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$,the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ contains all elements of $A$,$B$,and $C$.
Therefore,$\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ is a universal set for $A$,$B$,and $C$.

(B) universal set $U$ must contain all elements of the given sets $A$,$B$,and $C$.
Given sets are $A = \{1, 3, 5\}$,$B = \{2, 4, 6\}$,and $C = \{0, 2, 4, 6, 8\}$.
The union of these sets is $A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$.
For a set to be a universal set,it must contain all elements of $A \cup B \cup C$.
Comparing the options,the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ contains all elements of $A$,$B$,and $C$.
Therefore,option $B$ is the correct universal set.

(N/A) $A = \{ x : x \text{ is a natural number and a multiple of } 3 \} = \{ 3, 6, 9, 12, \ldots \}$
$B = \{ x : x \text{ is a natural number less than } 6 \} = \{ 1, 2, 3, 4, 5 \}$
$A \cup B = \{ 1, 2, 3, 4, 5, 6, 9, 12, \ldots \}$
$\therefore A \cup B = \{ x : x = 1, 2, 4, 5 \text{ or } x \text{ is a multiple of } 3 \}$

(A) The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$,in $B$,or in both.
Given sets are $A = \{1, 2, 3\}$ and $B = \varnothing$ (the empty set).
Therefore,$A \cup B = \{1, 2, 3\} \cup \varnothing = \{1, 2, 3\}$.
Thus,$A \cup B = A$.

(D) Given sets are:
$A = \{1, 2, 3, 4, 5, \ldots \}$
$D = \{2, 3, 5, 7, 11, \ldots \}$
Since every prime number is a natural number,$D \subset A$.
Therefore,the intersection of $A$ and $D$ is the set of all prime numbers.
$A \cap D = \{x : x \text{ is a prime number}\} = D$.

(A) Two sets are disjoint if their intersection is the empty set,denoted by $\varnothing$.
Let $A = \{ x : x \text{ is an even integer} \}$ and $B = \{ x : x \text{ is an odd integer} \}$.
The intersection of $A$ and $B$ is given by:
$A \cap B = \{ x : x \text{ is an even integer and } x \text{ is an odd integer} \}$.
Since no integer can be both even and odd,$A \cap B = \varnothing$.
Therefore,the pair of sets $A$ and $B$ is disjoint.

(A) The difference of two sets $X$ and $Y$,denoted by $X - Y$,is the set of elements which are in $X$ but not in $Y$.
Given $X = \{a, b, c, d\}$ and $Y = \{f, b, d, g\}$.
The elements of $X$ are $a, b, c, d$.
The elements of $Y$ are $f, b, d, g$.
The elements present in $X$ but not in $Y$ are $a$ and $c$.
Therefore,$X - Y = \{a, c\}$.

(B) $R$ is the set of real numbers.
$Q$ is the set of rational numbers.
The set of real numbers $R$ is the union of the set of rational numbers $Q$ and the set of irrational numbers $I$.
Thus,$R = Q \cup I$.
Therefore,$R - Q = I$,which is the set of irrational numbers.

(A) True.
Two sets are disjoint if their intersection is the empty set,denoted by $\varnothing$.
Let $A = \{2, 6, 10\}$ and $B = \{3, 7, 11\}$.
The intersection $A \cap B = \{2, 6, 10\} \cap \{3, 7, 11\} = \varnothing$.
Since the intersection is empty,the sets are disjoint.

(N/A) Let $X$ be the set of letters in "$CATARACT$".
Then $X = \{C, A, T, R\}$.
Let $Y$ be the set of letters in "$TRACT$".
Then $Y = \{T, R, A, C, T\} = \{T, R, A, C\}$.
Since every element in $X$ is in $Y$ and every element in $Y$ is in $X$,it follows that $X = Y$.

(N/A) Let $A = \{-1, 0, 1\}$.
The number of elements in $A$ is $n = 3$. The total number of subsets is given by $2^n = 2^3 = 8$.
The subset of $A$ having $0$ elements is the empty set $\phi$.
The subsets of $A$ having $1$ element are $\{-1\}, \{0\}, \{1\}$.
The subsets of $A$ having $2$ elements are $\{-1, 0\}, \{-1, 1\}, \{0, 1\}$.
The subset of $A$ having $3$ elements is $\{-1, 0, 1\}$.
Thus,all the subsets of $A$ are $\phi, \{-1\}, \{0\}, \{1\}, \{-1, 0\}, \{-1, 1\}, \{0, 1\}, \{-1, 0, 1\}$.

Let $x \in A$. Since $A \subseteq A \cup B$,we have $x \in A \cup B$.
Given $A \cup B = A \cap B$,it follows that $x \in A \cap B$.
By definition of intersection,$x \in A$ and $x \in B$.
Thus,$x \in B$,which implies $A \subseteq B$.
Similarly,let $y \in B$. Since $B \subseteq A \cup B$,we have $y \in A \cup B$.
Given $A \cup B = A \cap B$,it follows that $y \in A \cap B$.
By definition of intersection,$y \in A$ and $y \in B$.
Thus,$y \in A$,which implies $B \subseteq A$.
Since $A \subseteq B$ and $B \subseteq A$,we conclude $A = B$.

(N/A) Let $X \in P(A \cap B).$ Then $X \subset A \cap B.$
So,$X \subset A$ and $X \subset B.$
Therefore,$X \in P(A)$ and $X \in P(B),$ which implies $X \in P(A) \cap P(B).$
This gives $P(A \cap B) \subset P(A) \cap P(B).$
Let $Y \in P(A) \cap P(B).$ Then $Y \in P(A)$ and $Y \in P(B).$
So,$Y \subset A$ and $Y \subset B.$
Therefore,$Y \subset A \cap B,$ which implies $Y \in P(A \cap B).$
This gives $P(A) \cap P(B) \subset P(A \cap B).$
Hence,$P(A \cap B) = P(A) \cap P(B).$

(D) Given $A = \{ x: x \in \mathbb{R} \text{ and } x^2 - 8x + 12 = 0 \}$.
Solving the quadratic equation $x^2 - 8x + 12 = 0$:
$(x - 2)(x - 6) = 0$,which gives $x = 2$ or $x = 6$.
Thus,$A = \{2, 6\}$.
Given sets are $B = \{2, 4, 6\}$,$C = \{2, 4, 6, 8, \dots\}$,and $D = \{6\}$.
Comparing the elements:
$D = \{6\} \subset A = \{2, 6\}$.
$A = \{2, 6\} \subset B = \{2, 4, 6\}$.
$B = \{2, 4, 6\} \subset C = \{2, 4, 6, 8, \dots\}$.
Therefore,the subset relations are $D \subset A \subset B \subset C$.

(N/A) The statement is False.
To disprove the statement,we provide a counterexample.
Let $A = \{2\}$,$B = \{0, 2\}$,and $C = \{1, \{0, 2\}, 3\}$.
Here,$A \subset B$ because every element of $A$ (which is $2$) is also an element of $B$.
Also,$B \in C$ because the set $B = \{0, 2\}$ is an element of $C$.
However,$A \notin C$ because the set $A = \{2\}$ is not one of the elements of $C$ (the elements of $C$ are $1$,$\{0, 2\}$,and $3$).
Thus,the statement is false.

(A) The statement is True.
Proof:
Let $A \subset B$ and $B \subset C$.
Let $x$ be an arbitrary element such that $x \in A$.
Since $A \subset B$,by the definition of a subset,$x \in B$.
Since $B \subset C$,by the definition of a subset,$x \in C$.
Since $x \in A$ implies $x \in C$,it follows that $A \subset C$.

(N/A) The statement is False.
To disprove the statement,we need to find a counterexample where $A \not\subset B$ and $B \not\subset C$ are true,but $A \subset C$ is also true.
Let $A = \{1, 2\}$,$B = \{3, 4\}$,and $C = \{1, 2, 5\}$.
Here,$A \not\subset B$ (since $1 \in A$ but $1 \notin B$) and $B \not\subset C$ (since $3 \in B$ but $3 \notin C$).
However,$A \subset C$ because every element of $A$ is also an element of $C$.
Thus,the statement is false.

(N/A) The statement is $False$.
To disprove the statement,we provide a counterexample.
Let $A = \{1, 2\}$ and $B = \{1, 3\}$.
Here,$A \not\subset B$ because $2 \in A$ but $2 \notin B$.
Let $x = 2$. Then $x \in A$ and $A \not\subset B$ are satisfied.
However,$x \notin B$ because $2 \notin \{1, 3\}$.
Thus,the statement is $False$.

(A) The statement is True.
Given: $A \subset B$ and $x \notin B$.
To prove: $x \notin A$.
Proof by contradiction:
Suppose $x \in A$.
Since $A \subset B$,every element of $A$ must be an element of $B$.
Therefore,$x \in A$ implies $x \in B$.
However,we are given that $x \notin B$.
This is a contradiction.
Therefore,our assumption that $x \in A$ is false.
Hence,$x \notin A$.

(N/A) Given: $A \cup B = A \cup C$ and $A \cap B = A \cap C$.
To prove: $B = C$.
Let $x \in B$.
Since $B \subseteq A \cup B$,we have $x \in A \cup B$.
Given $A \cup B = A \cup C$,it follows that $x \in A \cup C$.
This implies $x \in A$ or $x \in C$.
Case $I$: If $x \in A$,then since $x \in B$,we have $x \in A \cap B$.
Since $A \cap B = A \cap C$,it follows that $x \in A \cap C$,which implies $x \in C$.
Case $II$: If $x \in C$,then $x \in C$ is already satisfied.
In both cases,$x \in C$. Thus,$B \subseteq C$.
Similarly,by taking $x \in C$,we can show $C \subseteq B$.
Since $B \subseteq C$ and $C \subseteq B$,we conclude $B = C$.

(A) We need to show that $(i) \Leftrightarrow (ii) \Leftrightarrow (iii) \Leftrightarrow (iv)$.
$(i) \Leftrightarrow (ii)$:
Assume $A \subset B$. If $x \in A - B$,then $x \in A$ and $x \notin B$. But $A \subset B$ implies $x \in A \Rightarrow x \in B$,which contradicts $x \notin B$. Thus,$A - B = \phi$.
Conversely,assume $A - B = \phi$. If $x \in A$,then $x$ must be in $B$ (otherwise $x \in A - B$,which is empty). Thus,$A \subset B$.
$(i) \Leftrightarrow (iii)$:
Assume $A \subset B$. Since $B \subset A \cup B$ is always true,we only need to show $A \cup B \subset B$. If $x \in A \cup B$,then $x \in A$ or $x \in B$. If $x \in A$,then $x \in B$ (as $A \subset B$). If $x \in B$,then $x \in B$. Thus,$A \cup B \subset B$,so $A \cup B = B$.
Conversely,assume $A \cup B = B$. If $x \in A$,then $x \in A \cup B$. Since $A \cup B = B$,$x \in B$. Thus,$A \subset B$.
$(i) \Leftrightarrow (iv)$:
Assume $A \subset B$. Since $A \cap B \subset A$ is always true,we show $A \subset A \cap B$. If $x \in A$,then $x \in B$ (as $A \subset B$). Thus,$x \in A$ and $x \in B$,so $x \in A \cap B$. Thus,$A \cap B = A$.
Conversely,assume $A \cap B = A$. If $x \in A$,then $x \in A \cap B$,which implies $x \in A$ and $x \in B$. Thus,$x \in B$,so $A \subset B$.

(N/A) Let $A \subset B$.
To show: $(C - B) \subset (C - A)$.
Let $x \in (C - B)$.
$\Rightarrow x \in C$ and $x \notin B$.
Since $A \subset B$,if $x \notin B$,then $x \notin A$.
$\Rightarrow x \in C$ and $x \notin A$.
$\Rightarrow x \in (C - A)$.
Therefore,$(C - B) \subset (C - A)$.

Given that $P(A) = P(B)$.
To show: $A = B$.
Let $x \in A$.
Since $A \in P(A)$ and $P(A) = P(B)$,it follows that $A \in P(B)$.
By the definition of the power set,if $A \in P(B)$,then $A \subseteq B$.
Therefore,$x \in A \implies x \in B$,which means $A \subseteq B$.
Similarly,let $y \in B$. Since $B \in P(B)$ and $P(B) = P(A)$,it follows that $B \in P(A)$.
By the definition of the power set,if $B \in P(A)$,then $B \subseteq A$.
Therefore,$y \in B \implies y \in A$,which means $B \subseteq A$.
Since $A \subseteq B$ and $B \subseteq A$,we conclude that $A = B$.

(N/A) The statement is $False$.
Let $A = \{0, 1\}$ and $B = \{1, 2\}$.
Then $A \cup B = \{0, 1, 2\}$.
The power set $P(A) = \{\varnothing, \{0\}, \{1\}, \{0, 1\}\}$.
The power set $P(B) = \{\varnothing, \{1\}, \{2\}, \{1, 2\}\}$.
Therefore,$P(A) \cup P(B) = \{\varnothing, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{1, 2\}\}$.
The power set $P(A \cup B) = \{\varnothing, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{0, 2\}, \{1, 2\}, \{0, 1, 2\}\}$.
Comparing the two sets,we observe that $\{0, 2\} \in P(A \cup B)$ but $\{0, 2\} \notin P(A) \cup P(B)$.
Thus,$P(A) \cup P(B) \neq P(A \cup B)$.

(N/A) To prove: $A \cap (A \cup B) = A$
Using the distributive law of intersection over union:
$A \cap (A \cup B) = (A \cap A) \cup (A \cap B)$
Since $A \cap A = A$ (idempotent law):
$= A \cup (A \cap B)$
Since $(A \cap B) \subseteq A$,the union of $A$ and a subset of $A$ is $A$ itself (absorption law):
$= A$

(N/A) Let $A = \{0, 1\}, B = \{0, 2, 3\},$ and $C = \{0, 4, 5\}$.
Accordingly,$A \cap B = \{0\}$ and $A \cap C = \{0\}$.
Here,$A \cap B = A \cap C = \{0\}$.
However,$B \neq C$ because $2 \in B$ and $2 \notin C$.

Given that $A \cap X = \phi$,$B \cap X = \phi$,and $A \cup X = B \cup X$ for some set $X$.
To show: $A = B$.
We know that $A = A \cap (A \cup X)$.
Substituting $A \cup X = B \cup X$,we get:
$A = A \cap (B \cup X)$.
Using the Distributive law,$A = (A \cap B) \cup (A \cap X)$.
Since $A \cap X = \phi$,we have $A = (A \cap B) \cup \phi = A \cap B$ ... $(1)$.
Similarly,$B = B \cap (B \cup X)$.
Substituting $B \cup X = A \cup X$,we get:
$B = B \cap (A \cup X)$.
Using the Distributive law,$B = (B \cap A) \cup (B \cap X)$.
Since $B \cap X = \phi$,we have $B = (B \cap A) \cup \phi = B \cap A = A \cap B$ ... $(2)$.
From $(1)$ and $(2)$,we conclude that $A = B$.

(N/A) Let $A = \{0, 1\}, B = \{1, 2\},$ and $C = \{2, 0\}$.
Accordingly,$A \cap B = \{1\}, B \cap C = \{2\},$ and $A \cap C = \{0\}$.
Therefore,$A \cap B, B \cap C,$ and $A \cap C$ are non-empty sets.
However,$A \cap B \cap C = \varnothing$.

When two dice are thrown,the sample space $S$ contains $36$ outcomes.
$B$ is the event of getting an odd number on the first die:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
$C$ is the event of getting a sum $\leq 5$:
$C = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$
The event $B$ or $C$ is the union $B \cup C$,which includes all elements present in either $B$ or $C$:
$B \cup C = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$

(B) The sample space $S$ consists of $36$ outcomes.
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Since $B$ is the event of getting an odd number on the first die,its complement $B^{\prime}$ is the event of getting an even number on the first die.
Thus,$B^{\prime} = A$.
Now,$A \cap B^{\prime} = A \cap A = A$.
Since $A$ is not an empty set $(A \neq \phi)$,$A \cap B^{\prime} \neq \phi$.
Therefore,$A$ and $B^{\prime}$ are not mutually exclusive.
Thus,the given statement is false.

(C) The number of subsets of a set with $k$ elements is $2^k$.
Given that the number of subsets of $A$ is $112$ more than the number of subsets of $B$,we have the equation: $2^m - 2^n = 112$.
We can rewrite this as $2^n(2^{m-n} - 1) = 112$.
Since $112 = 16 \times 7 = 2^4 \times (2^3 - 1)$,we compare the terms:
$2^n = 2^4 \implies n = 4$.
$2^{m-n} - 1 = 2^3 - 1 \implies m - n = 3$.
Substituting $n = 4$,we get $m - 4 = 3$,so $m = 7$.
Therefore,$m \times n = 7 \times 4 = 28$.

(A) Given the equation: $(1+a^2)(1+b^2) = 4ab$
Expanding the left side: $1 + b^2 + a^2 + a^2b^2 = 4ab$
Rearranging the terms: $a^2 - 2ab + b^2 + a^2b^2 - 2ab + 1 = 0$
This can be written as: $(a-b)^2 + (ab-1)^2 = 0$
Since $a$ and $b$ are real numbers,the sum of squares is zero if and only if each term is zero:
$(a-b)^2 = 0 \implies a = b$
$(ab-1)^2 = 0 \implies ab = 1$
Substituting $b=a$ into $ab=1$,we get $a^2 = 1$,which implies $a = 1$ or $a = -1$.
However,the problem states $a > 0$ and $a \neq 1$. Thus,there is no value of $b$ that satisfies the equation under the given constraints.
Therefore,the set $S$ is an empty set.

(D) The set $S_n$ contains $18$ consecutive integers.
$(a)$ If $n=10$,$S_{10} = \{11, 12, \ldots, 28\}$. The multiples of $19$ in this range is $19$. However,if $n=19$,$S_{19} = \{20, 21, \ldots, 37\}$. The multiples of $19$ are $38$,which is not in $S_{19}$. Thus,$(a)$ is false.
$(b)$ $S_n$ always contains a prime number (by Bertrand's Postulate,there is always a prime between $k$ and $2k-2$ for $k > 3$). This is true,but we must check if $(d)$ is a stronger or more specific property often tested in this context. Let's re-evaluate $(d)$.
$(c)$ For $n=10$,$S_{10} = \{11, 12, \ldots, 28\}$. The multiples of $5$ are $15, 20, 25$. There are only $3$ multiples. Thus,$(c)$ is false.
$(d)$ In any set of $18$ consecutive integers,there are at most $6$ primes. For example,in the range $[11, 28]$,the primes are $11, 13, 17, 19, 23$. There are $5$ primes. As $n$ increases,the density of primes decreases. Thus,$S_n$ has at most $6$ primes. This is a known property for this specific set size. Therefore,$(d)$ is the correct statement.

(C) The set $A = S \times S$ contains $n^2$ elements.
$A$ subset $B$ of $A$ is a good subset if it contains all elements of the form $(x, x)$ for $x \in S$.
There are $n$ such elements: $(1, 1), (2, 2), \ldots, (n, n)$.
For $B$ to be a good subset,these $n$ elements must be present in $B$.
The remaining elements in $A$ are those $(a, b)$ where $a \neq b$. The number of such elements is $n^2 - n = n(n - 1)$.
Each of these $n(n - 1)$ elements can either be in $B$ or not in $B$.
Therefore,the number of ways to choose the remaining elements is $2^{n(n - 1)}$.
Thus,the total number of good subsets is $2^{n(n - 1)}$.

(A) Let $n$ be an integer. When a nonzero digit $d$ is appended to the right of $n$,the new number is $10n + d$.
We are given that $10n + d$ is divisible by $d$ for all $d \in \{1, 2, \dots, 9\}$.
This implies $\frac{10n + d}{d} = \frac{10n}{d} + 1$ must be an integer for all $d \in \{1, 2, \dots, 9\}$.
Thus,$10n$ must be divisible by every $d \in \{1, 2, \dots, 9\}$.
This means $10n$ must be a multiple of the least common multiple of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
$\text{LCM}(1, 2, 3, 4, 5, 6, 7, 8, 9) = 2^3 \times 3^2 \times 5 \times 7 = 2520$.
So,$10n$ must be a multiple of $2520$,which means $n$ must be a multiple of $252$.
The smallest positive integer $n$ is $252$.
The product of the digits of $N = 252$ is $2 \times 5 \times 2 = 20$.

(A) Given the equation $x^2 - 2^y = 2023$ where $x, y \in \mathbb{N}$.
Rearranging the equation,we get $x^2 - 2023 = 2^y$.
If $y = 1$,then $x^2 - 2023 = 2^1 = 2$,which implies $x^2 = 2025$,so $x = 45$.
If $y = 2$,then $x^2 - 2023 = 4$,which implies $x^2 = 2027$ (not a perfect square).
If $y = 3$,then $x^2 - 2023 = 8$,which implies $x^2 = 2031$ (not a perfect square).
If $y = 4$,then $x^2 - 2023 = 16$,which implies $x^2 = 2039$ (not a perfect square).
If $y = 5$,then $x^2 - 2023 = 32$,which implies $x^2 = 2055$ (not a perfect square).
If $y = 6$,then $x^2 - 2023 = 64$,which implies $x^2 = 2087$ (not a perfect square).
If $y = 7$,then $x^2 - 2023 = 128$,which implies $x^2 = 2151$ (not a perfect square).
If $y = 8$,then $x^2 - 2023 = 256$,which implies $x^2 = 2279$ (not a perfect square).
If $y = 9$,then $x^2 - 2023 = 512$,which implies $x^2 = 2535$ (not a perfect square).
If $y = 10$,then $x^2 - 2023 = 1024$,which implies $x^2 = 3047$ (not a perfect square).
For $y > 1$,considering the equation modulo $3$: $x^2 - 2^y \equiv 2023 \pmod 3 \Rightarrow x^2 - (-1)^y \equiv 1 \pmod 3$.
If $y$ is even,$x^2 - 1 \equiv 1 \Rightarrow x^2 \equiv 2 \pmod 3$,which is impossible.
If $y$ is odd,$x^2 + 1 \equiv 1 \Rightarrow x^2 \equiv 0 \pmod 3$,so $x$ must be a multiple of $3$. Let $x = 3k$.
Then $9k^2 - 2^y = 2023$. For $y > 1$,$2^y$ is a multiple of $4$. $9k^2 - 2023 = 2^y$. This leads to no further solutions.
Thus,the only solution is $(45, 1)$.
The sum $\sum_{(x, y) \in C} (x + y) = 45 + 1 = 46$.

(A) We are given $A = \{1, 2, \ldots, 10\}$ and $B = \left\{\frac{m}{n} : m, n \in A, m < n, \gcd(m, n) = 1\right\}$.
To find $n(B)$,we count the number of fractions $\frac{m}{n}$ for each $n \in \{2, 3, \ldots, 10\}$ such that $m < n$ and $\gcd(m, n) = 1$.
For $n=2$: $m \in \{1\}$,$\gcd(1, 2) = 1$. Count = $1$.
For $n=3$: $m \in \{1, 2\}$,$\gcd(1, 3) = 1, \gcd(2, 3) = 1$. Count = $2$.
For $n=4$: $m \in \{1, 3\}$,$\gcd(1, 4) = 1, \gcd(3, 4) = 1$. Count = $2$.
For $n=5$: $m \in \{1, 2, 3, 4\}$,all are coprime to $5$. Count = $4$.
For $n=6$: $m \in \{1, 5\}$,$\gcd(1, 6) = 1, \gcd(5, 6) = 1$. Count = $2$.
For $n=7$: $m \in \{1, 2, 3, 4, 5, 6\}$,all are coprime to $7$. Count = $6$.
For $n=8$: $m \in \{1, 3, 5, 7\}$,all are coprime to $8$. Count = $4$.
For $n=9$: $m \in \{1, 2, 4, 5, 7, 8\}$,all are coprime to $9$. Count = $6$.
For $n=10$: $m \in \{1, 3, 7, 9\}$,all are coprime to $10$. Count = $4$.
Summing these counts: $1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 + 4 = 31$.
Thus,$n(B) = 31$.