Mix Examples of Set Theory Questions in English

(B) Given that $A \cup B = A \cup C$ and $A \cap B = A \cap C$.
Consider the set $B$. We can write $B = B \cap (A \cup B)$.
Since $A \cup B = A \cup C$,we have $B = B \cap (A \cup C)$.
Using the distributive law,$B = (B \cap A) \cup (B \cap C)$.
Since $A \cap B = A \cap C$,we have $B = (A \cap C) \cup (B \cap C)$.
By the distributive law,$B = (A \cup B) \cap C$.
Since $A \cup B = A \cup C$,we have $B = (A \cup C) \cap C$.
Since $(A \cup C) \cap C = C$,it follows that $B = C$.

(B) Let $S = \sum_{k=0}^{99} \left[ \frac{1}{2} + \frac{k}{100} \right]$.
For $0 \le k \le 49$,we have $0 \le \frac{k}{100} \le 0.49$,so $\frac{1}{2} + \frac{k}{100} = 0.5 + \frac{k}{100} < 1$. Thus,$\left[ \frac{1}{2} + \frac{k}{100} \right] = 0$ for $k = 0, 1, \dots, 49$.
For $50 \le k \le 99$,we have $0.50 \le \frac{k}{100} \le 0.99$,so $1 \le \frac{1}{2} + \frac{k}{100} < 1.49$. Thus,$\left[ \frac{1}{2} + \frac{k}{100} \right] = 1$ for $k = 50, 51, \dots, 99$.
The number of terms from $k = 50$ to $k = 99$ is $99 - 50 + 1 = 50$.
Therefore,the sum is $0 \times 50 + 1 \times 50 = 50$.

(B) Given $A \cap B = A \cap C$ and $A \cup B = A \cup C$.
Consider $B = B \cap (A \cup B)$.
Since $A \cup B = A \cup C$,we have $B = B \cap (A \cup C)$.
Using the distributive law,$B = (B \cap A) \cup (B \cap C)$.
Since $A \cap B = A \cap C$,we have $B = (A \cap C) \cup (B \cap C)$.
$B = (A \cup B) \cap C$.
Since $A \cup B = A \cup C$,we have $B = (A \cup C) \cap C$.
Since $(A \cup C) \cap C = C$,it follows that $B = C$.

(B) Let $t = \sqrt{x}$,where $t \ge 0$. The equation becomes $2|t - 3| + t(t - 6) + 6 = 0$.
Case $I$: $0 \le t < 3$ (i.e.,$0 \le x < 9$)
$2(3 - t) + t^2 - 6t + 6 = 0$
$6 - 2t + t^2 - 6t + 6 = 0$
$t^2 - 8t + 12 = 0$
$(t - 6)(t - 2) = 0$
$t = 6$ or $t = 2$.
Since $0 \le t < 3$,we have $t = 2$,which implies $x = 4$.
Case $II$: $t \ge 3$ (i.e.,$x \ge 9$)
$2(t - 3) + t^2 - 6t + 6 = 0$
$2t - 6 + t^2 - 6t + 6 = 0$
$t^2 - 4t = 0$
$t(t - 4) = 0$
$t = 0$ or $t = 4$.
Since $t \ge 3$,we have $t = 4$,which implies $x = 16$.
Thus,$S = \{4, 16\}$,which contains exactly two elements.

(B) Given that $A \cup B = A \cup C$ and $A \cap B = A \cap C$.
Consider the set $B$. We can write $B = B \cap (A \cup B)$.
Since $A \cup B = A \cup C$,we have $B = B \cap (A \cup C)$.
Using the distributive law,$B = (B \cap A) \cup (B \cap C)$.
Since $A \cap B = A \cap C$,we have $B = (A \cap C) \cup (B \cap C)$.
Now consider the set $C$. We can write $C = C \cap (A \cup C)$.
Since $A \cup C = A \cup B$,we have $C = C \cap (A \cup B)$.
Using the distributive law,$C = (C \cap A) \cup (C \cap B)$.
Since $A \cap C = A \cap B$,we have $C = (A \cap B) \cup (C \cap B)$.
Comparing the expressions for $B$ and $C$,we see that $B = (A \cap C) \cup (B \cap C)$ and $C = (A \cap B) \cup (C \cap B)$.
Since $A \cap C = A \cap B$,it follows that $B = C$.

(B) Let $S = (A - B) \cup (B - C) \cup (C - A)$.
We know that $A - B = A \cap B^c$,$B - C = B \cap C^c$,and $C - A = C \cap A^c$.
So,$S = (A \cap B^c) \cup (B \cap C^c) \cup (C \cap A^c)$.
By De Morgan's Law and set properties,the complement $S'$ is given by:
$S' = (A \cap B^c)' \cap (B \cap C^c)' \cap (C \cap A^c)'$
$S' = (A^c \cup B) \cap (B^c \cup C) \cap (C^c \cup A)$.
Expanding this expression:
$(A^c \cup B) \cap (B^c \cup C) = (A^c \cap B^c) \cup (A^c \cap C) \cup (B \cap B^c) \cup (B \cap C) = (A^c \cap B^c) \cup (A^c \cap C) \cup (B \cap C)$.
Multiplying by $(C^c \cup A)$ results in $A \cap B \cap C$.

(A) For the roots of the quadratic equation $x^{2} - (m+1)x + m+4 = 0$ to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = (m+1)^{2} - 4(m+4) \geq 0$
$m^{2} + 2m + 1 - 4m - 16 \geq 0$
$m^{2} - 2m - 15 \geq 0$
$(m-5)(m+3) \geq 0$
Thus,$m \in (-\infty, -3] \cup [5, \infty)$,so $A = (-\infty, -3] \cup [5, \infty)$.
Given $B = [-3, 5)$.
Now,evaluate the options:
$A - B = (-\infty, -3) \cup [5, \infty)$ (True)
$A \cap B = \{-3\}$ (True)
$B - A = (-3, 5)$ (True)
$A \cup B = (-\infty, -3] \cup [-3, 5) \cup [5, \infty) = (-\infty, \infty) = R$ (True)
Wait,checking the options again: $A - B = (-\infty, -3) \cup [5, \infty)$. Since $-3 \in B$,it is removed from $A$. So $A - B = (-\infty, -3) \cup [5, \infty)$. This is true.
All options provided are actually true. However,if we re-examine $A \cup B$,it is $R$. If we re-examine $B - A$,it is $(-3, 5)$.
Given the standard nature of this problem,there might be a typo in the question's options. Assuming the question asks for the false statement,and all are true,we identify that $A \cup B = R$ is true,$A \cap B = \{-3\}$ is true,$B - A = (-3, 5)$ is true,and $A - B = (-\infty, -3) \cup [5, \infty)$ is true. Since all are true,the question is flawed.

(B) Given the equation: $x^3 - [x]^3 = (x - [x])^3$.
Let ${x} = x - [x]$. Then the equation becomes $x^3 - [x]^3 = {x}^3$.
We know that $x^3 - [x]^3 = (x - [x])(x^2 + x[x] + [x]^2)$.
So,$(x - [x])(x^2 + x[x] + [x]^2) = (x - [x])^3$.
This implies $(x - [x])[(x^2 + x[x] + [x]^2) - (x - [x])^2] = 0$.
$(x - [x])[x^2 + x[x] + [x]^2 - (x^2 - 2x[x] + [x]^2)] = 0$.
$(x - [x])[3x[x]] = 0$.
This gives two cases:
Case $1$: $x - [x] = 0 \implies x \in \mathbb{Z}$.
Case $2$: $3x[x] = 0 \implies x = 0$ or $[x] = 0$.
If $[x] = 0$,then $0 \le x < 1$.
Combining these,$A = \mathbb{Z} \cup [0, 1)$.
Since $A$ contains the interval $[0, 1)$ but also includes isolated points like $\dots, -2, -1, 2, 3, \dots$,it is not an interval.
Thus,$A$ contains an interval,but is not an interval.

(A) Let $n = 2018$. We are given $N = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
For inequality $I$,we use the Cauchy-Schwarz inequality in the form $(\sum a_k b_k)^2 \leq (\sum a_k^2)(\sum b_k^2)$.
Let $a_k = \sqrt{k}$ and $b_k = \sqrt{k} x_k$. Then:
$\left(\sum_{k=1}^{n} \sqrt{k} \cdot \sqrt{k} x_k\right)^2 \leq \left(\sum_{k=1}^{n} (\sqrt{k})^2\right) \left(\sum_{k=1}^{n} (\sqrt{k} x_k)^2\right)$
$\left(\sum_{k=1}^{n} k x_k\right)^2 \leq \left(\sum_{k=1}^{n} k\right) \left(\sum_{k=1}^{n} k x_k^2\right) = N \sum_{k=1}^{n} k x_k^2$.
Thus,$I$ is true.
For inequality $II$,we note that since $k \geq 1$,we have $k^2 \geq k$ for all $k \in \{1, 2, \dots, n\}$.
Therefore,$\sum_{k=1}^{n} k^2 x_k^2 \geq \sum_{k=1}^{n} k x_k^2$.
Since $I$ is true,we have $\left(\sum_{k=1}^{n} k x_k\right)^2 \leq N \sum_{k=1}^{n} k x_k^2 \leq N \sum_{k=1}^{n} k^2 x_k^2$.
Thus,$II$ is also true.
Therefore,both $I$ and $II$ are true.

(A) For set $A$,we have $\cos^2(\sin \theta) + \sin^2(\cos \theta) = 1$.
Using the identity $\sin^2 x + \cos^2 x = 1$,this equation holds if and only if $\sin^2 \theta = \cos^2 \theta$,which implies $\tan^2 \theta = 1$,so $\tan \theta = \pm 1$.
Thus,$\theta = n\pi \pm \frac{\pi}{4}$.
For set $B$,we have $\cos(\sin \theta) \sin(\cos \theta) = 0$.
This implies $\cos(\sin \theta) = 0$ or $\sin(\cos \theta) = 0$.
Since $-1 \le \sin \theta \le 1$,$\cos(\sin \theta)$ can never be $0$ because $\cos x = 0$ at $x = \pm \frac{\pi}{2} \approx \pm 1.57$,which is outside the range $[-1, 1]$.
Similarly,$\sin(\cos \theta) = 0$ implies $\cos \theta = n\pi$. Since $-1 \le \cos \theta \le 1$,the only possible value is $\cos \theta = 0$.
Thus,$\theta = (2n+1)\frac{\pi}{2}$.
Comparing the sets,$A = \{n\pi \pm \frac{\pi}{4}\}$ and $B = \{(2n+1)\frac{\pi}{2}\}$,we see that there is no common value.
Therefore,$A \cap B = \emptyset$.

(D) two-digit number $\overline{ab}$ is almost prime if it can be transformed into a prime number by changing at most one digit.
Every two-digit number $\overline{ab}$ can be written as $10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.
There are $90$ two-digit numbers in total,ranging from $10$ to $99$.
Any prime number $p$ (where $10 \le p \le 99$) is itself an almost prime number because we can change zero digits to obtain a prime.
For any non-prime two-digit number,we check if it can be changed into a prime by altering one digit.
Since there are many primes ending in $1, 3, 7, 9$,almost every two-digit number can be converted into a prime by changing the units digit $b$.
For example,any number $\overline{a0}, \overline{a2}, \overline{a4}, \overline{a5}, \overline{a6}, \overline{a8}$ can be changed to a prime by modifying the last digit.
Exhaustive checking shows that all $90$ two-digit numbers satisfy this condition.
Thus,the total count is $90$.

(A) Let $p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ where $a_i \in \mathbb{Z}$.
Given $p(2) = 2$ and $p(4) = 5$.
By the property of polynomials with integer coefficients,for any two distinct integers $a$ and $b$,$(a-b)$ must divide $(p(a) - p(b))$.
Here,$a = 4$ and $b = 2$.
So,$(4-2)$ must divide $(p(4) - p(2))$.
$(4-2) = 2$ and $(p(4) - p(2)) = 5 - 2 = 3$.
Since $2$ does not divide $3$,there is no such polynomial with integer coefficients.
Therefore,the number of such polynomials is $0$.

(B) Given $\text{HCF}(x, y) = 16$ and $\text{LCM}(x, y) = 48000$.
Let $x = 16a$ and $y = 16b$,where $\text{HCF}(a, b) = 1$.
We know that $\text{LCM}(x, y) = \text{HCF}(x, y) \times a \times b$.
$48000 = 16 \times a \times b
\implies ab = \frac{48000}{16} = 3000$.
Prime factorization of $3000 = 3 \times 1000 = 3^1 \times 2^3 \times 5^3$.
Since $\text{HCF}(a, b) = 1$,the prime factors $2^3, 3^1, 5^3$ must be distributed between $a$ and $b$ such that no prime factor is common to both.
For each prime factor $p^k$,we have two choices: either it is a factor of $a$ or it is a factor of $b$.
There are $3$ distinct prime factors $(2, 3, 5)$.
For each prime factor,there are $2$ choices.
Total number of pairs $(a, b) = 2^3 = 8$.
Since each pair $(a, b)$ corresponds to a unique pair $(x, y)$,the number of elements in $S$ is $8$.

(B) For set $A$: $[x + 3] + [x + 4] \leq 3 \implies [x] + 3 + [x] + 4 \leq 3$.
$2[x] + 7 \leq 3 \implies 2[x] \leq -4 \implies [x] \leq -2$.
Since $[x] \leq -2$,we have $x < -1$,so $A = (-\infty, -1)$.
For set $B$: The sum is a geometric series $\sum_{n=1}^{\infty} \frac{3}{10^n} = 3 \left( \frac{1/10}{1 - 1/10} \right) = 3 \left( \frac{1/10}{9/10} \right) = 3 \left( \frac{1}{9} \right) = \frac{1}{3} = 3^{-1}$.
The inequality becomes $3^x (3^{-1})^{x-3} < 3^{-3x}$.
$3^x \cdot 3^{-x+3} < 3^{-3x} \implies 3^3 < 3^{-3x}$.
$27 < 3^{-3x} \implies 3^3 < 3^{-3x}$.
Comparing exponents: $3 < -3x \implies x < -1$.
Thus,$B = (-\infty, -1)$.
Since $A = (-\infty, -1)$ and $B = (-\infty, -1)$,we conclude $A = B$.

(D) For each element in the set $S$,there are $3$ possibilities regarding its membership in two disjoint subsets $A$ and $B$:
$1$. The element is in $A$ but not in $B$.
$2$. The element is in $B$ but not in $A$.
$3$. The element is in neither $A$ nor $B$.
Since there are $n = 4$ elements,the total number of ordered pairs of disjoint subsets $(A, B)$ is $3^n = 3^4 = 81$.
To find the number of unordered pairs ${A, B}$,we must account for the case where $A = B$. Since $A$ and $B$ must be disjoint,$A = B$ implies $A = B = \emptyset$. This occurs in exactly $1$ case.
For all other cases where $A \neq B$,the pair ${A, B}$ is counted twice in the ordered list (as $(A, B)$ and $(B, A)$).
The number of unordered pairs is given by $\frac{3^n + 1}{2} = \frac{3^4 + 1}{2} = \frac{81 + 1}{2} = \frac{82}{2} = 41$.

(D) We are given the sets $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + y| \geq 3\}$ and $B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x| + |y| \leq 3\}$.
We need to find the set $C = \{(x, y) \in A \cap B : x = 0 \text{ or } y = 0\}$.
Case $1$: If $x = 0$,then $(0, y) \in A \cap B$.
From $B$,$|0| + |y| \leq 3 \implies |y| \leq 3 \implies -3 \leq y \leq 3$.
From $A$,$|0 + y| \geq 3 \implies |y| \geq 3$.
Combining these,$|y| = 3$,so $y = 3$ or $y = -3$. Thus,$(0, 3)$ and $(0, -3)$ are in $C$.
Case $2$: If $y = 0$,then $(x, 0) \in A \cap B$.
From $B$,$|x| + |0| \leq 3 \implies |x| \leq 3 \implies -3 \leq x \leq 3$.
From $A$,$|x + 0| \geq 3 \implies |x| \geq 3$.
Combining these,$|x| = 3$,so $x = 3$ or $x = -3$. Thus,$(3, 0)$ and $(-3, 0)$ are in $C$.
Therefore,$C = \{(3, 0), (-3, 0), (0, 3), (0, -3)\}$.
Now,we calculate $\sum_{(x, y) \in C} |x + y| = |3 + 0| + |-3 + 0| + |0 + 3| + |0 - 3| = 3 + 3 + 3 + 3 = 12$.

(C) For set $A$: $\log_{(2/\pi)}|\sin x| + \log_{(2/\pi)}|\cos x| = 2$
$\Rightarrow \log_{(2/\pi)}|\sin x \cos x| = 2$
$\Rightarrow |\sin x \cos x| = (2/\pi)^2 = 4/\pi^2$
$\Rightarrow |\frac{1}{2} \sin 2x| = 4/\pi^2$
$\Rightarrow |\sin 2x| = 8/\pi^2$
Since $8/\pi^2 \approx 8/9.86 < 1$,the equation $|\sin 2x| = 8/\pi^2$ has $4$ solutions in $(0, \pi) - \{\pi/2\}$.
For set $B$: Let $t = \sqrt{x} \geq 0$. The equation is $t(t - 4) - 3|t - 2| + 6 = 0$.
Case $1$: $0 \leq t < 2$. Then $|t - 2| = -(t - 2) = 2 - t$.
$t^2 - 4t - 3(2 - t) + 6 = 0$ $\Rightarrow t^2 - 4t - 6 + 3t + 6 = 0$ $\Rightarrow t^2 - t = 0$ $\Rightarrow t(t - 1) = 0$.
Since $0 \leq t < 2$,$t = 0$ and $t = 1$ are solutions. Thus $x = 0^2 = 0$ and $x = 1^2 = 1$.
Case $2$: $t \geq 2$. Then $|t - 2| = t - 2$.
$t^2 - 4t - 3(t - 2) + 6 = 0$ $\Rightarrow t^2 - 4t - 3t + 6 + 6 = 0$ $\Rightarrow t^2 - 7t + 12 = 0$ $\Rightarrow (t - 3)(t - 4) = 0$.
Thus $t = 3$ and $t = 4$. So $x = 3^2 = 9$ and $x = 4^2 = 16$.
Set $B = \{0, 1, 9, 16\}$,so $n(B) = 4$.
Since sets $A$ and $B$ are disjoint,$n(A \cup B) = n(A) + n(B) = 4 + 4 = 8$.

(D) Let the given sum be $S = \sum_{k=0}^{99} \left[\frac{1}{2} + \frac{k}{100}\right]$.
We observe the value of each term $\left[\frac{1}{2} + \frac{k}{100}\right]$:
For $0 \le k \le 49$,we have $\frac{1}{2} \le \frac{1}{2} + \frac{k}{100} \le \frac{1}{2} + \frac{49}{100} = 0.99$. Since $0 \le 0.5 + \frac{k}{100} < 1$,the greatest integer part is $0$. There are $50$ such terms (from $k=0$ to $k=49$).
For $50 \le k \le 99$,we have $\frac{1}{2} + \frac{50}{100} \le \frac{1}{2} + \frac{k}{100} \le \frac{1}{2} + \frac{99}{100}$,which simplifies to $1 \le \frac{1}{2} + \frac{k}{100} \le 1.49$. Since $1 \le 0.5 + \frac{k}{100} < 2$,the greatest integer part is $1$. There are $50$ such terms (from $k=50$ to $k=99$).
Thus,$S = (50 \times 0) + (50 \times 1) = 50$.

(C) Given the values at $x = 0, 1, 2, 4$,we assume the function $f(x)$ is a polynomial of degree $3$ or less,implying the $4^{th}$ order forward difference is zero: $\Delta^{4} f(0) = 0$.
Using the operator $E$,where $Ef(x) = f(x+1)$,we have $(E-1)^{4} f(0) = 0$.
Expanding this,we get $(E^{4}-4E^{3}+6E^{2}-4E+1) f(0) = 0$.
This simplifies to $f(4) - 4f(3) + 6f(2) - 4f(1) + f(0) = 0$.
Substituting the known values $f(0)=1, f(1)=3, f(2)=9, f(4)=81$:
$81 - 4f(3) + 6(9) - 4(3) + 1 = 0$.
$81 - 4f(3) + 54 - 12 + 1 = 0$.
$124 - 4f(3) = 0$.
$4f(3) = 124$.
$f(3) = 31$.

(C) Let the degree of each vertex be $K$.
Since the graph is regular,every vertex has the same degree $K$.
The sum of the degrees of all vertices in a graph is given by the product of the number of vertices and the degree of each vertex.
Given that there are $15$ vertices and the sum of their degrees is $60$,we have:
$15 \times K = 60$
Dividing both sides by $15$,we get:
$K = \frac{60}{15} = 4$
Therefore,the degree of each vertex is $4$.

(B) Given $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$.
Notice the pattern in the values: $P(x) = x^2 + 1$ for $x = 0, 1, 2, 3, 4$.
Let $Q(x) = P(x) - (x^2 + 1)$.
Since $P(x)$ is a polynomial of degree $5$ and $x^2 + 1$ is of degree $2$,$Q(x)$ is a polynomial of degree $5$ with roots $0, 1, 2, 3, 4$.
Thus,$Q(x) = k(x)(x-1)(x-2)(x-3)(x-4)$ for some constant $k$.
Comparing the coefficient of $x^5$ in $P(x)$,we see $k = 1$.
So,$P(x) = x(x-1)(x-2)(x-3)(x-4) + x^2 + 1$.
To find $P(5)$,substitute $x = 5$:
$P(5) = 5(5-1)(5-2)(5-3)(5-4) + 5^2 + 1$
$P(5) = 5 \times 4 \times 3 \times 2 \times 1 + 25 + 1$
$P(5) = 120 + 26 = 146$.

(C) For each element in the set $A$,there are $3$ possibilities:
$1$. The element is in subset $P$.
$2$. The element is in subset $Q$.
$3$. The element is in neither $P$ nor $Q$.
Since the subsets $P$ and $Q$ must be mutually disjoint,an element cannot be in both $P$ and $Q$.
Given that set $A$ has $n = 5$ elements,each of the $5$ elements has $3$ choices.
Therefore,the total number of ways to select the subsets $P$ and $Q$ is $3^n = 3^5 = 243$.

(D) Let $n(A) = m$ and $n(B) = n$.
Given that $P_B$ has $112$ elements more than $P_A$,we have $n(P_B) - n(P_A) = 112$.
Since $n(P_A) = 2^m$ and $n(P_B) = 2^n$,we get $2^n - 2^m = 112$.
Factoring out $2^m$,we have $2^m(2^{n-m} - 1) = 112$.
We can write $112$ as $16 \times 7 = 2^4 \times (8 - 1) = 2^4(2^3 - 1)$.
Comparing both sides,we get $m = 4$ and $n - m = 3$,which implies $n = 7$.
The number of injective functions from $A$ to $B$ is given by the formula $^n P_m$.
Substituting the values,we get $^7 P_4 = \frac{7!}{(7-4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.

(B) For set $A$,we require $x^2-8x+15 \geq 0$.
$(x-3)(x-5) \geq 0$,which gives $x \in (-\infty, 3] \cup [5, \infty)$.
For set $B$,we solve $\frac{x-3}{2x-5} - \frac{x-6}{2x-11} < 0$.
$\frac{(x-3)(2x-11) - (x-6)(2x-5)}{(2x-5)(2x-11)} < 0$.
$\frac{(2x^2-17x+33) - (2x^2-17x+30)}{(2x-5)(2x-11)} < 0$.
$\frac{3}{(2x-5)(2x-11)} < 0$.
This implies $(2x-5)(2x-11) < 0$,so $x \in \left(\frac{5}{2}, \frac{11}{2}\right)$.
Finally,$A \cap B = ((-\infty, 3] \cup [5, \infty)) \cap \left(\frac{5}{2}, \frac{11}{2}\right) = \left(\frac{5}{2}, 3\right] \cup \left[5, \frac{11}{2}\right)$.

(A) For $p$: $6^m + 9^n \equiv 1^m + (-1)^n \equiv 1 + (-1)^n \pmod{5}$.
For this to be a multiple of $5$,$1 + (-1)^n \equiv 0 \pmod{5}$,which implies $(-1)^n = -1$.
This occurs when $n$ is odd. In the set $\{1, 2, \ldots, 50\}$,there are $25$ odd values for $n$ and $50$ values for $m$.
Thus,$p = 50 \times 25 = 1250$.
For $q$: $m + n$ must be a square of a prime number. The possible squares of primes are $2^2 = 4$,$3^2 = 9$,$5^2 = 25$,and $7^2 = 49$.

$m + n = 4$	$3$ pairs: $(1,3), (2,2), (3,1)$
$m + n = 9$	$8$ pairs: $(1,8), \ldots, (8,1)$
$m + n = 25$	$24$ pairs: $(1,24), \ldots, (24,1)$
$m + n = 49$	$48$ pairs: $(1,48), \ldots, (48,1)$

$q = 3 + 8 + 24 + 48 = 83$.
$p + q = 1250 + 83 = 1333$.

(B) The set $A$ consists of terms $a_n = 1 + (n-1)5 = 5n - 4$ for $n = 1, 2, \dots, 101$. The maximum value is $5(101) - 4 = 501$.
The set $B$ consists of terms $b_m = 9 + (m-1)7 = 7m + 2$ for $m = 1, 2, \dots, 71$. The maximum value is $7(71) + 2 = 499$.
For an element $x$ to be in $A \cap B$,we have $x = 5n - 4 = 7m + 2$,which implies $5n = 7m + 6$.
Testing values for $m$: If $m=1, x=9$ (not in $A$). If $m=2, x=16$ ($16 = 5(4)-4$,so $16 \in A$).
The common terms form an $A$.$P$. with common difference $\text{lcm}(5, 7) = 35$. Thus,$x = 35k + 16$.
We require $16 \le 35k + 16 \le 499$,which gives $0 \le k \le 13.8$,so $k \in \{0, 1, 2, \dots, 13\}$.
We want $x$ to be divisible by $3$: $35k + 16 \equiv 2k + 1 \equiv 0 \pmod 3$.
This implies $2k \equiv -1 \equiv 2 \pmod 3$,so $k \equiv 1 \pmod 3$.
The possible values for $k$ are $1, 4, 7, 10, 13$.
There are $5$ such values.