Multinomial theorem, Number of divisors Questions in English

(C) To find the number of divisors of $9600$,we first find its prime factorization.
$9600 = 96 \times 100 = (32 \times 3) \times (4 \times 25) = (2^5 \times 3) \times (2^2 \times 5^2) = 2^7 \times 3^1 \times 5^2$.
If a number $N$ is expressed as $N = p_1^{a} \times p_2^{b} \times p_3^{c}$,then the total number of divisors is given by $(a + 1)(b + 1)(c + 1)$.
Here,$a = 7$,$b = 1$,and $c = 2$.
Therefore,the number of divisors $= (7 + 1)(1 + 1)(2 + 1) = 8 \times 2 \times 3 = 48$.

(D) The given number is $N = a^1 b^2 c^2 d^1 e^1$.
Since $a, b, c, d, e$ are prime integers,the total number of divisors of $N$ is given by the product of (exponent + $1$) for each prime factor.
Total number of divisors $= (1 + 1)(2 + 1)(2 + 1)(1 + 1)(1 + 1) = 2 \times 3 \times 3 \times 2 \times 2 = 72$.
We need to find the number of divisors excluding $1$ as a factor.
Number of divisors excluding $1 = 72 - 1 = 71$.

(A) The prime factorization of $960$ is $960 = 2^6 \times 3^1 \times 5^1.$
The sum of all positive divisors of a number $N = p_1^{a_1} \times p_2^{a_2} \times p_3^{a_3}$ is given by the formula $\left( \frac{p_1^{a_1+1}-1}{p_1-1} \right) \left( \frac{p_2^{a_2+1}-1}{p_2-1} \right) \left( \frac{p_3^{a_3+1}-1}{p_3-1} \right).$
Substituting the values $p_1=2, a_1=6, p_2=3, a_2=1, p_3=5, a_3=1$:
Sum $= \left( \frac{2^{6+1}-1}{2-1} \right) \times \left( \frac{3^{1+1}-1}{3-1} \right) \times \left( \frac{5^{1+1}-1}{5-1} \right)$
Sum $= \left( \frac{128-1}{1} \right) \times \left( \frac{9-1}{2} \right) \times \left( \frac{25-1}{4} \right)$
Sum $= 127 \times \frac{8}{2} \times \frac{24}{4}$
Sum $= 127 \times 4 \times 6 = 3048$.

(C) To find the exponent of a prime $p$ in $n!$,we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.
Here,$n = 100$ and $p = 3$.
$E_3(100!) = \lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{3^2} \rfloor + \lfloor \frac{100}{3^3} \rfloor + \lfloor \frac{100}{3^4} \rfloor$
$E_3(100!) = \lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{9} \rfloor + \lfloor \frac{100}{27} \rfloor + \lfloor \frac{100}{81} \rfloor$
$E_3(100!) = 33 + 11 + 3 + 1 = 48$.
Thus,the exponent of $3$ in $100!$ is $48$.

(A) Let $N = 111...1$ ($91$ times).
$N = \sum_{k=0}^{90} 10^k = \frac{10^{91}-1}{10-1} = \frac{10^{91}-1}{9}$.
Since $91 = 7 \times 13$,we can write $10^{91}-1 = (10^7)^{13}-1$.
Using the algebraic identity $x^n-1 = (x-1)(x^{n-1} + x^{n-2} + ... + 1)$,we have:
$10^{91}-1 = (10^7-1)((10^7)^{12} + (10^7)^{11} + ... + 1)$.
Thus,$N = \frac{10^7-1}{9} \times ((10^7)^{12} + (10^7)^{11} + ... + 1)$.
Since $\frac{10^7-1}{9} = 1111111$ and the second factor is clearly greater than $1$,$N$ is a product of two integers greater than $1$.
Therefore,$N$ is a composite number,which means it is not a prime number.

(A) According to Fermat's Little Theorem,if $p$ is a prime number,then for any integer $n$,$n^p \equiv n \pmod{p}$.
This implies that $n^p - n$ is divisible by $p$ for any integer $n$.
Since the options provided focus on the set of natural numbers,the statement holds true for any natural number $n \geq 1$.
For example,let $n = 2$ and $p = 3$:
$2^3 - 2 = 8 - 2 = 6$,which is divisible by $3$.

(A) First,find the prime factorization of $n = 38808$:
$38808 = 8 \times 4851 = 2^3 \times 9 \times 539 = 2^3 \times 3^2 \times 7^2 \times 11^1$.
The total number of divisors is calculated using the formula $(a+1)(b+1)(c+1)(d+1)$ where $a, b, c, d$ are the exponents of the prime factors:
Total divisors $= (3+1)(2+1)(2+1)(1+1) = 4 \times 3 \times 3 \times 2 = 72$.
Since the question asks for the number of divisors excluding $1$ and $n$,we subtract $2$ from the total:
Required number of divisors $= 72 - 2 = 70$.

(A) To find the highest power of a prime $p$ in $n!$,we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.
Here,$n = 100$ and $p = 3$.
$E_3(100!) = \lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{9} \rfloor + \lfloor \frac{100}{27} \rfloor + \lfloor \frac{100}{81} \rfloor$.
$E_3(100!) = 33 + 11 + 3 + 1 = 48$.

(A) First,find the prime factorization of $9000$:
$9000 = 9 \times 1000 = 3^2 \times 10^3 = 3^2 \times (2 \times 5)^3 = 2^3 \times 3^2 \times 5^3$.
The sum of divisors of a number $N = p_1^{a} \times p_2^{b} \times p_3^{c}$ is given by the formula:
$S = \frac{p_1^{a+1}-1}{p_1-1} \times \frac{p_2^{b+1}-1}{p_2-1} \times \frac{p_3^{c+1}-1}{p_3-1}$.
For $9000 = 2^3 \times 3^2 \times 5^3$:
Sum $= (2^0 + 2^1 + 2^2 + 2^3) \times (3^0 + 3^1 + 3^2) \times (5^0 + 5^1 + 5^2 + 5^3)$.
Sum $= (1 + 2 + 4 + 8) \times (1 + 3 + 9) \times (1 + 5 + 25 + 125)$.
Sum $= (15) \times (13) \times (156)$.
Note: The provided option $A$ is $16 \times 13 \times 156$. Re-evaluating the sum of powers of $2$: $(2^0 + 2^1 + 2^2 + 2^3) = 1+2+4+8 = 15$.
Given the options,there is a slight discrepancy in the first term ($16$ vs $15$). However,$15 \times 13 \times 156 = 30420$. The sum of divisors is $30420$.

(C) To find the number of divisors of $9600$,we first find its prime factorization.
$9600 = 96 \times 100 = (32 \times 3) \times (10^2) = (2^5 \times 3^1) \times (2^2 \times 5^2) = 2^7 \times 3^1 \times 5^2$.
If a number $N$ is expressed as $p_1^{a} \times p_2^{b} \times p_3^{c}$,the total number of divisors is given by $(a+1)(b+1)(c+1)$.
Here,$a=7$,$b=1$,and $c=2$.
Total number of divisors = $(7+1)(1+1)(2+1) = 8 \times 2 \times 3 = 48$.

(A) First,find the prime factorization of $38808$:
$38808 = 8 \times 4851 = 8 \times 9 \times 539 = 8 \times 9 \times 7 \times 77 = 8 \times 9 \times 7 \times 7 \times 11 = 2^3 \times 3^2 \times 7^2 \times 11^1$.
The total number of divisors is given by the product of (exponent + $1$) for each prime factor:
Total divisors = $(3+1)(2+1)(2+1)(1+1) = 4 \times 3 \times 3 \times 2 = 72$.
Since we must exclude $1$ and $n$ (the number itself),we subtract $2$ from the total count:
Number of divisors excluding $1$ and $n$ = $72 - 2 = 70$.

(B) The number of terms in the expansion of a multinomial $(a_1 + a_2 + \dots + a_k)^n$ is given by the formula $\binom{n+k-1}{k-1}$.
For the expression $(1 - \frac{2}{x} + \frac{4}{x^2})^n$,we have $k=3$ terms.
Thus,the number of terms is $\binom{n+3-1}{3-1} = \binom{n+2}{2} = \frac{(n+2)(n+1)}{2}$.
Given $\frac{(n+2)(n+1)}{2} = 28$,we have $(n+2)(n+1) = 56$.
Since $8 \times 7 = 56$,we get $n+2 = 8$,which implies $n = 6$.
The sum of the coefficients of an expansion is found by substituting the variable with $1$.
Sum of coefficients $= (1 - 2 + 4)^n = (3)^n$.
For $n = 6$,the sum is $3^6 = 729$.

(D) To find the unit digit of $843^{843}$,we look at the unit digit of $3^{843}$. The powers of $3$ follow a cycle of $4$: $3^1=3, 3^2=9, 3^3=27, 3^4=81$.
$843 \div 4$ leaves a remainder of $3$,so the unit digit of $3^{843}$ is the same as $3^3$,which is $7$.
To find the unit digit of $492^{295}$,we look at the unit digit of $2^{295}$. The powers of $2$ follow a cycle of $4$: $2^1=2, 2^2=4, 2^3=8, 2^4=16$.
$295 \div 4$ leaves a remainder of $3$,so the unit digit of $2^{295}$ is the same as $2^3$,which is $8$.
The unit digit of the sum is the unit digit of $(7 + 8) = 15$,which is $5$.

(C) To find the highest power of a prime $p$ dividing $n!$,we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.
For $n=33$ and $p=2$,the exponent of $2$ in $33!$ is:
$E_2(33!) = \lfloor \frac{33}{2} \rfloor + \lfloor \frac{33}{4} \rfloor + \lfloor \frac{33}{8} \rfloor + \lfloor \frac{33}{16} \rfloor + \lfloor \frac{33}{32} \rfloor$
$E_2(33!) = 16 + 8 + 4 + 2 + 1 = 31$.
Thus,$33!$ is divisible by $2^n$ for all $n \in \{1, 2, 3, \ldots, 31\}$.
The sum of all possible values of $n$ is the sum of the first $31$ natural numbers:
$S = \frac{31(31+1)}{2} = \frac{31 \times 32}{2} = 31 \times 16 = 496$.

(C) The number of ordered pairs $(x, y)$ such that $xy = n$ where $n$ is a factor of $72$ is equivalent to finding the number of pairs $(x, y)$ such that $xy$ divides $72$.
Let $72 = 2^3 \times 3^2$.
We want to find the number of pairs $(x, y)$ such that $xy = 2^a 3^b$ where $0 \le a \le 3$ and $0 \le b \le 2$.
Let $x = 2^{a_1} 3^{b_1}$ and $y = 2^{a_2} 3^{b_2}$.
Then $xy = 2^{a_1+a_2} 3^{b_1+b_2} = 2^a 3^b$.
For a fixed $n = 2^a 3^b$,the number of solutions to $a_1 + a_2 = a$ is $(a+1)$ and $b_1 + b_2 = b$ is $(b+1)$.
Total pairs = $\sum_{a=0}^{3} \sum_{b=0}^{2} (a+1)(b+1) = (1+2+3+4)(1+2+3) = 10 \times 6 = 60$.

(D) The expression $N = n(n+1)(n+2)(n+3)$ is the product of four consecutive integers.
Among any four consecutive integers,one is divisible by $4$,another is divisible by $2$,and at least one is divisible by $3$.
Thus,$N$ is always divisible by $4 \times 2 \times 3 = 24$.
We can rewrite $N$ as:
$N = [n(n+3)] \times [(n+1)(n+2)]$
$N = (n^2 + 3n)(n^2 + 3n + 2)$
Let $x = n^2 + 3n$. Then $N = x(x+2) = x^2 + 2x = (x+1)^2 - 1$.
Since $N = (n^2 + 3n + 1)^2 - 1$,$N$ is one less than a perfect square.
For any natural number $n \ge 1$,$N > 0$. Since $N$ is not a perfect square,the number of divisors $d$ of $N$ must be even (because divisors of a non-perfect square always come in pairs $(a, b)$ such that $ab = N$ and $a \neq b$).

(A) Given the equation $xyz = 2^5 \times 3^2 \times 5^2$.
Let $x = 2^{a_1} 3^{b_1} 5^{c_1}$,$y = 2^{a_2} 3^{b_2} 5^{c_2}$,and $z = 2^{a_3} 3^{b_3} 5^{c_3}$,where $a_i, b_i, c_i \ge 0$.
Since $x, y, z$ are natural numbers,$a_i, b_i, c_i$ must be non-negative integers.
The exponents must satisfy:
$a_1 + a_2 + a_3 = 5$
$b_1 + b_2 + b_3 = 2$
$c_1 + c_2 + c_3 = 2$
Using the stars and bars formula,the number of non-negative integer solutions for $x_1 + x_2 + \dots + x_k = n$ is given by $\binom{n+k-1}{k-1}$.
For $a_1 + a_2 + a_3 = 5$: $\binom{5+3-1}{3-1} = \binom{7}{2} = 21$.
For $b_1 + b_2 + b_3 = 2$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.
For $c_1 + c_2 + c_3 = 2$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.
The total number of solutions is $21 \times 6 \times 6 = 756$.

(B) Any divisor of $2^5 \times 3^4 \times 5^2$ is of the form $2^a \times 3^b \times 5^c$ where $0 \le a \le 5$,$0 \le b \le 4$,and $0 \le c \le 2$.
The sum of the divisors is given by the product of the geometric series for each prime factor:
Sum $= (1 + 2 + 2^2 + 2^3 + 2^4 + 2^5) \times (1 + 3 + 3^2 + 3^3 + 3^4) \times (1 + 5 + 5^2)$
Using the formula for the sum of a geometric series $\frac{r^n - 1}{r - 1}$:
Sum $= \left( \frac{2^6 - 1}{2 - 1} \right) \times \left( \frac{3^5 - 1}{3 - 1} \right) \times \left( \frac{5^3 - 1}{5 - 1} \right)$
Sum $= (64 - 1) \times \left( \frac{243 - 1}{2} \right) \times \left( \frac{125 - 1}{4} \right)$
Sum $= 63 \times \frac{242}{2} \times \frac{124}{4}$
Sum $= 63 \times 121 \times 31$
Prime factorization of $63 = 3^2 \times 7$.
So,Sum $= 3^2 \times 7^1 \times 11^2 \times 31$.

(A) First,find the prime factorization of $480$:
$480 = 48 \times 10 = (16 \times 3) \times (2 \times 5) = 2^4 \times 3^1 \times 2^1 \times 5^1 = 2^5 \times 3^1 \times 5^1$.
Any divisor of $480$ is of the form $d = 2^a \times 3^b \times 5^c$,where $0 \leq a \leq 5$,$0 \leq b \leq 1$,and $0 \leq c \leq 1$.
We want the divisor to be of the form $4n + 2$,which means $d$ must be divisible by $2$ but not by $4$.
This implies $a$ must be exactly $1$ (so that $d = 2^1 \times 3^b \times 5^c = 2(3^b \times 5^c)$).
For $b$,we have $2$ choices ($0$ or $1$).
For $c$,we have $2$ choices ($0$ or $1$).
Total number of such divisors = $1 \times 2 \times 2 = 4$.
The divisors are:
$2^1 \times 3^0 \times 5^0 = 2$
$2^1 \times 3^1 \times 5^0 = 6$
$2^1 \times 3^0 \times 5^1 = 10$
$2^1 \times 3^1 \times 5^1 = 30$
All these are of the form $4n + 2$ $(2=4(0)+2, 6=4(1)+2, 10=4(2)+2, 30=4(7)+2)$.

(B) The given expression is $(1 + x^n + x^{253})^{10}$.
Using the multinomial theorem,the general term is given by $\frac{10!}{a!b!c!} (1)^a (x^n)^b (x^{253})^c$,where $a + b + c = 10$.
We need the coefficient of $x^{1012}$,so we set the exponent of $x$ to $1012$:
$nb + 253c = 1012$.
Since $c \leq 10$ and $253 \times 4 = 1012$,we test values for $c$:
If $c = 4$,then $nb = 1012 - 253(4) = 0$. Since $n$ is a positive integer,$b$ must be $0$.
Then $a = 10 - b - c = 10 - 0 - 4 = 6$.
The coefficient is $\frac{10!}{6!0!4!} = ^{10}C_4$.
If $c < 4$,then $nb = 253(4-c)$. For $c=3$,$nb = 253$,which is impossible for $n \leq 22$ and $b \leq 10$ (as $22 \times 10 = 220 < 253$).
Thus,the only solution is $a=6, b=0, c=4$.
The coefficient is $^{10}C_4$.

(D) Given $y+z=5$ and $\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$.
From the second equation,$\frac{y+z}{yz} = \frac{5}{6}$.
Substituting $y+z=5$,we get $\frac{5}{yz} = \frac{5}{6}$,which implies $yz = 6$.
We have $y+z=5$ and $yz=6$. The quadratic equation $t^2 - 5t + 6 = 0$ has roots $t=2$ and $t=3$.
Since $y > z$,we have $y=3$ and $z=2$.
The number $n = 2^{x} \cdot 3^{3} \cdot 5^{2}$.
An odd divisor of $n$ must be of the form $3^{a} \cdot 5^{b}$,where $0 \le a \le 3$ and $0 \le b \le 2$.
The number of choices for $a$ is $(3+1) = 4$.
The number of choices for $b$ is $(2+1) = 3$.
Total number of odd divisors $= 4 \times 3 = 12$.

(A) Let $N$ be a $4$-digit number such that $\gcd(N, 18) = 3$.
Since $\gcd(N, 18) = 3$,$N$ must be a multiple of $3$ but not a multiple of $2$ (because $18 = 2 \times 3^2$,and if $N$ were even,$\gcd(N, 18)$ would be at least $6$) and not a multiple of $9$ (because if $N$ were a multiple of $9$,$\gcd(N, 18)$ would be $9$ or $18$).
Thus,$N$ must be an odd multiple of $3$ that is not divisible by $9$.
First,find the number of $4$-digit odd multiples of $3$:
The smallest $4$-digit odd multiple of $3$ is $1005$ and the largest is $9999$.
These form an arithmetic progression: $1005, 1011, \dots, 9999$.
The number of terms is $\frac{9999 - 1005}{6} + 1 = \frac{8994}{6} + 1 = 1499 + 1 = 1500$.
Next,find the number of $4$-digit odd multiples of $9$:
The smallest $4$-digit odd multiple of $9$ is $1017$ and the largest is $9999$.
These form an arithmetic progression: $1017, 1035, \dots, 9999$.
The number of terms is $\frac{9999 - 1017}{18} + 1 = \frac{8982}{18} + 1 = 499 + 1 = 500$.
The number of such $N$ is $1500 - 500 = 1000$.

(A) The given number is $N = (10)^{10} \cdot (11)^{11} \cdot (13)^{13} = 2^{10} \cdot 5^{10} \cdot 11^{11} \cdot 13^{13}$.
$A$ divisor $d$ of $N$ is of the form $2^a \cdot 5^b \cdot 11^c \cdot 13^d$,where $0 \le a \le 10, 0 \le b \le 10, 0 \le c \le 11, 0 \le d \le 13$.
For $d$ to be of the form $4n+1$,$d$ must be odd,so $a=0$.
Thus,$d = 5^b \cdot 11^c \cdot 13^d$.
Modulo $4$,we have $5 \equiv 1 \pmod{4}$,$11 \equiv -1 \equiv 3 \pmod{4}$,and $13 \equiv 1 \pmod{4}$.
So,$d \equiv 1^b \cdot (-1)^c \cdot 1^d \equiv (-1)^c \pmod{4}$.
For $d \equiv 1 \pmod{4}$,we must have $(-1)^c = 1$,which means $c$ must be even.
Possible values for $c \in \{0, 2, 4, 6, 8, 10\}$,which gives $6$ choices.
Possible values for $b \in \{0, 1, 2, \dots, 10\}$,which gives $11$ choices.
Possible values for $d \in \{0, 1, 2, \dots, 13\}$,which gives $14$ choices.
The total number of such divisors is $11 \times 6 \times 14 = 924$.

(D) The general term in the multinomial expansion is given by:
$T_{r_1, r_2, r_3} = \frac{10!}{r_1! r_2! r_3!} (3x^3)^{r_1} (-2x^2)^{r_2} (5x^{-5})^{r_3}$
$T_{r_1, r_2, r_3} = \frac{10!}{r_1! r_2! r_3!} (3)^{r_1} (-2)^{r_2} (5)^{r_3} x^{3r_1 + 2r_2 - 5r_3}$
For the constant term,the power of $x$ must be $0$:
$3r_1 + 2r_2 - 5r_3 = 0$ $(1)$
Also,$r_1 + r_2 + r_3 = 10$ $(2)$
From $(2)$,$r_2 = 10 - r_1 - r_3$. Substituting into $(1)$:
$3r_1 + 2(10 - r_1 - r_3) - 5r_3 = 0$
$3r_1 + 20 - 2r_1 - 2r_3 - 5r_3 = 0$
$r_1 + 20 = 7r_3$
Testing integer values for $r_3$:
If $r_3 = 3$,then $r_1 = 7(3) - 20 = 1$. Then $r_2 = 10 - 1 - 3 = 6$.
Constant term $= \frac{10!}{1! 6! 3!} (3)^1 (-2)^6 (5)^3$
$= \frac{10 \times 9 \times 8 \times 7}{3 \times 2 \times 1} \times 3 \times 64 \times 125$
$= (10 \times 3 \times 4 \times 7) \times 3 \times 2^6 \times 5^3$
$= (840) \times 3 \times 2^6 \times 125 = (2^3 \times 3 \times 5 \times 7) \times 3 \times 2^6 \times 5^3$
$= 2^9 \times 3^2 \times 5^4 \times 7^1 = 2^9 \cdot l$,where $l = 3^2 \cdot 5^4 \cdot 7^1$ is an odd integer.
Thus,$k = 9$.

(B) Let the number of games won by the three players be $x, y,$ and $z$ respectively.
We are given that the total number of games is $x + y + z = 9$.
The total score for each player must be zero at the end.
For a player who wins $x$ games and loses the remaining $(9-x)$ games,the total score is $2x - 1(9-x) = 3x - 9$.
Setting the score to zero,we get $3x - 9 = 0$,which implies $x = 3$.
Similarly,$y = 3$ and $z = 3$.
Thus,each player must win exactly $3$ games out of $9$.
The number of ways to distribute the wins among the three players is given by the multinomial coefficient:
$\frac{9!}{3!3!3!} = \frac{362880}{6 \times 6 \times 6} = \frac{362880}{216} = 1680$.

(A) Let the number be $N$. When a non-zero digit $c$ is appended to $N$,the new number is $10N + c$.
Given that $10N + c$ is divisible by $c$ for all $c \in \{1, 2, \dots, 9\}$.
This implies that $10N$ must be divisible by $c$ for all $c \in \{1, 2, \dots, 9\}$.
Therefore,$10N$ must be a multiple of the least common multiple $(LCM)$ of the digits $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
$LCM(1, 2, 3, 4, 5, 6, 7, 8, 9) = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520$.
So,$10N$ must be a multiple of $2520$,which means $N$ must be a multiple of $252$.
The least positive integer $N$ is $252$.
The sum of the digits of $N$ is $2 + 5 + 2 = 9$.

(B) fraction $\frac{1}{n}$ has a terminating decimal expansion if and only if the prime factorization of the denominator $n$ is of the form $2^a \times 5^b$,where $a, b \ge 0$ and $a+b > 0$.
We need to find the number of integers $n \in \{2, 3, \ldots, 200\}$ of the form $2^a \times 5^b$.
Possible values for $n$ are:
- $2^a$: $2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64, 2^7=128$
- $5^b$: $5^1=5, 5^2=25, 5^3=125$
- $2^a \times 5^b$:
- $b=1$: $2^1 \times 5^1=10, 2^2 \times 5^1=20, 2^3 \times 5^1=40, 2^4 \times 5^1=80, 2^5 \times 5^1=160$
- $b=2$: $2^1 \times 5^2=50, 2^2 \times 5^2=100, 2^3 \times 5^2=200$
Listing all unique values: $2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128, 160, 200$.
Counting these,we get a total of $18$ values.

(A) The number of integers less than $n$ and coprime to $n$ is given by Euler's totient function $\phi(n)$.
Given $n = p_1 \times p_2$,where $p_1$ and $p_2$ are distinct primes,the formula for $\phi(n)$ is $\phi(n) = (p_1-1)(p_2-1)$.
We are given $\phi(43361) = 42900$.
So,$(p_1-1)(p_2-1) = 42900$.
Expanding this,$p_1 p_2 - p_1 - p_2 + 1 = 42900$.
Since $p_1 p_2 = 43361$,we have $43361 - (p_1+p_2) + 1 = 42900$.
$43362 - (p_1+p_2) = 42900$.
$p_1+p_2 = 43362 - 42900 = 462$.
Thus,the sum of the two prime factors is $462$.

(B) To find the largest $k$ such that $24^k$ divides $13!$,we first find the prime factorization of $13!$.
$13! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$.
We have $24^k = (2^3 \times 3)^k = 2^{3k} \times 3^k$.
For $24^k$ to divide $13!$,we must have $3k \le 10$ and $k \le 5$.
From $3k \le 10$,we get $k \le \frac{10}{3} \approx 3.33$.
From $k \le 5$,we get $k \le 5$.
The largest integer $k$ satisfying both conditions is $k = 3$.

(C) The $6$-digit number is of the form $ababab$.
$ababab = 10^5 a + 10^4 b + 10^3 a + 10^2 b + 10a + b$
$= (10^5 + 10^3 + 10)a + (10^4 + 10^2 + 1)b$
$= (101010)a + (10101)b = 10101(10a + b)$
$= (3 \times 7 \times 13 \times 37)(10a + b)$
For the number to be a product of exactly $6$ distinct primes,$(10a + b)$ must be a product of $2$ distinct primes,and these primes must not be in the set $\{3, 7, 13, 37\}$.
The possible values for $(10a + b)$ where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$ are numbers from $10$ to $99$.
We check products of two distinct primes $p_1 \times p_2$ such that $p_1, p_2 \notin \{3, 7, 13, 37\}$:
$10 = 2 \times 5$
$14 = 2 \times 7$ (contains $7$,invalid)
$22 = 2 \times 11$
$26 = 2 \times 13$ (contains $13$,invalid)
$34 = 2 \times 17$
$38 = 2 \times 19$
$46 = 2 \times 23$
$55 = 5 \times 11$
$58 = 2 \times 29$
$62 = 2 \times 31$
$65 = 5 \times 13$ (contains $13$,invalid)
$74 = 2 \times 37$ (contains $37$,invalid)
$82 = 2 \times 41$
$85 = 5 \times 17$
$86 = 2 \times 43$
$91 = 7 \times 13$ (contains $7, 13$,invalid)
$94 = 2 \times 47$
$95 = 5 \times 19$
Counting the valid products: $10, 22, 34, 38, 46, 55, 58, 62, 82, 85, 86, 94, 95$. There are $13$ such numbers.

(A) Given the equation $\frac{m}{12} = \frac{12}{n}$.
Multiplying both sides,we get $mn = 144$.
The prime factorization of $144$ is $2^4 \times 3^2$.
The number of positive divisors of $144$ is $(4+1)(2+1) = 5 \times 3 = 15$.
Since $m$ and $n$ can be integers,they can be both positive or both negative.
If $m, n > 0$,there are $15$ possible pairs $(m, n)$.
If $m, n < 0$,there are also $15$ possible pairs $(m, n)$.
Therefore,the total number of ordered pairs $(m, n)$ is $15 + 15 = 30$.

(C) We are given that $(n+3)$ divides $(n^3-3)$.
Using polynomial division,we can write:
$\frac{n^3-3}{n+3} = \frac{n^3+27-30}{n+3} = \frac{(n+3)(n^2-3n+9)-30}{n+3} = (n^2-3n+9) - \frac{30}{n+3}$.
For the expression to be an integer,$(n+3)$ must be a divisor of $30$.
Since $n$ is a positive integer,$n \ge 1$,so $n+3 \ge 4$.
The divisors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$.
Considering the condition $n+3 \ge 4$,the possible values for $(n+3)$ are $5, 6, 10, 15, 30$.
Calculating $n$ for each case:
$n+3 = 5 \Rightarrow n = 2$
$n+3 = 6 \Rightarrow n = 3$
$n+3 = 10 \Rightarrow n = 7$
$n+3 = 15 \Rightarrow n = 12$
$n+3 = 30 \Rightarrow n = 27$
Thus,there are $5$ such positive integers $n$.

(B) For $\frac{1}{q}$ to have a terminating decimal expansion,$q$ must be of the form $2^m 5^n$ where $m, n \in \mathbb{W}$.
For $\sqrt{q}$ to be rational,$q$ must be a perfect square.
If $q = 2^m 5^n$ is a perfect square,then both $m$ and $n$ must be even.
Let $m = 2a$ and $n = 2b$ where $a, b \in \mathbb{W}$.
Then $q = 2^{2a} 5^{2b} = (2^a 5^b)^2$.
We are given $1 \leq q \leq 2021$,so $1 \leq (2^a 5^b)^2 \leq 2021$,which implies $1 \leq 2^a 5^b \leq \sqrt{2021} \approx 44.95$.
We list the possible values for $2^a 5^b \leq 44$:
If $b=0$: $2^a \leq 44 \Rightarrow a \in \{0, 1, 2, 3, 4, 5\}$ (Values: $1, 2, 4, 8, 16, 32$)
If $b=1$: $2^a \cdot 5 \leq 44$ $\Rightarrow 2^a \leq 8.8$ $\Rightarrow a \in \{0, 1, 2, 3\}$ (Values: $5, 10, 20, 40$)
If $b=2$: $2^a \cdot 25 \leq 44$ $\Rightarrow 2^a \leq 1.76$ $\Rightarrow a = 0$ (Value: $25$)
Total values for $2^a 5^b$ are $6 + 4 + 1 = 11$.
Thus,there are $11$ such integers $q$.

(B) The general term in the multinomial expansion is given by $\frac{5!}{n_1! n_2! n_3!} (2x)^{n_1} (x^{-7})^{n_2} (3x^2)^{n_3}$,where $n_1 + n_2 + n_3 = 5$.
This simplifies to $\frac{5!}{n_1! n_2! n_3!} 2^{n_1} 3^{n_3} x^{n_1 - 7n_2 + 2n_3}$.
For the constant term,the power of $x$ must be zero,so $n_1 - 7n_2 + 2n_3 = 0$.
Given $n_1 + n_2 + n_3 = 5$,we substitute $n_1 = 5 - n_2 - n_3$ into the equation:
$(5 - n_2 - n_3) - 7n_2 + 2n_3 = 0$ $\Rightarrow 5 - 8n_2 + n_3 = 0$ $\Rightarrow n_3 = 8n_2 - 5$.
If $n_2 = 1$,then $n_3 = 3$,which implies $n_1 = 5 - 1 - 3 = 1$.
Substituting these values into the coefficient formula:
$\frac{5!}{1! 1! 3!} (2)^1 (3)^3 = \frac{120}{6} \times 2 \times 27 = 20 \times 54 = 1080$.

(B) Given the function $f(x) = 2x^n + \lambda$.
Using the given values:
$f(4) = 2(4^n) + \lambda = 133$ --- $(1)$
$f(5) = 2(5^n) + \lambda = 255$ --- $(2)$
Subtracting equation $(1)$ from $(2)$:
$2(5^n - 4^n) = 255 - 133 = 122$
$5^n - 4^n = 61$
By testing values for $n \in N$:
For $n = 3$,$5^3 - 4^3 = 125 - 64 = 61$. Thus,$n = 3$.
Substituting $n = 3$ into equation $(1)$:
$2(4^3) + \lambda = 133$
$2(64) + \lambda = 133$
$128 + \lambda = 133 \Rightarrow \lambda = 5$.
Now,calculate $f(3) - f(2)$:
$f(3) = 2(3^3) + 5 = 2(27) + 5 = 59$
$f(2) = 2(2^3) + 5 = 2(8) + 5 = 21$
$f(3) - f(2) = 59 - 21 = 38$.
The divisors of $38$ are $1, 2, 19, 38$.
The sum of these divisors is $1 + 2 + 19 + 38 = 60$.

(A) Let $N$ be a $4$-digit number. We are given $\gcd(N, 54) = 2$.
Since $54 = 2 \times 3^3$,the condition $\gcd(N, 54) = 2$ implies that $N$ must be divisible by $2$ but not by $3$.
Total $4$-digit numbers range from $1000$ to $9999$,so there are $9000$ such numbers.
Number of $4$-digit numbers divisible by $2$ is $\lfloor \frac{9999}{2} \rfloor - \lfloor \frac{999}{2} \rfloor = 4999 - 499 = 4500$.
Number of $4$-digit numbers divisible by $6$ (i.e.,divisible by both $2$ and $3$) is $\lfloor \frac{9999}{6} \rfloor - \lfloor \frac{999}{6} \rfloor = 1666 - 166 = 1500$.
The number of $4$-digit numbers divisible by $2$ but not by $3$ is $4500 - 1500 = 3000$.

(B) To find the largest power of a prime $p$ that divides $n!$,we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \left[ \frac{n}{p^k} \right]$.
Here,$n = 66$ and $p = 3$.
$E_3(66!) = \left[ \frac{66}{3} \right] + \left[ \frac{66}{3^2} \right] + \left[ \frac{66}{3^3} \right]$
$E_3(66!) = \left[ \frac{66}{3} \right] + \left[ \frac{66}{9} \right] + \left[ \frac{66}{27} \right]$
$E_3(66!) = 22 + 7 + 2 = 31$.
Thus,the largest natural number $n$ is $31$.

(A) The general term in the expansion of $(1-x+2x^3)^{10}$ is given by $\frac{10!}{r_1! r_2! r_3!} (1)^{r_1} (-x)^{r_2} (2x^3)^{r_3} = \frac{10!}{r_1! r_2! r_3!} (-1)^{r_2} (2)^{r_3} x^{r_2+3r_3}$.
We require $r_1+r_2+r_3=10$ and $r_2+3r_3=7$,where $r_1, r_2, r_3 \ge 0$.
Possible non-negative integer solutions for $(r_1, r_2, r_3)$ are:
$1$. If $r_3=0$,then $r_2=7$,so $r_1=3$.
$2$. If $r_3=1$,then $r_2=4$,so $r_1=5$.
$3$. If $r_3=2$,then $r_2=1$,so $r_1=7$.
The coefficient is the sum of terms for these cases:
Case $1$: $\frac{10!}{3! 7! 0!} (-1)^7 (2)^0 = -120$.
Case $2$: $\frac{10!}{5! 4! 1!} (-1)^4 (2)^1 = 1260 \times 2 = 2520$.
Case $3$: $\frac{10!}{7! 1! 2!} (-1)^1 (2)^2 = 360 \times (-4) = -1440$.
Summing these: $-120 + 2520 - 1440 = 960$.

(C) The total number of $3$-digit numbers is $999 - 99 = 900$.
$A$ number is divisible by both $2$ and $3$ if it is divisible by $\text{lcm}(2, 3) = 6$.
The number of $3$-digit numbers divisible by $6$ is $\frac{900}{6} = 150$.
$A$ number is divisible by both $4$ and $9$ if it is divisible by $\text{lcm}(4, 9) = 36$.
The number of $3$-digit numbers divisible by $36$ is $\frac{900}{36} = 25$.
Since any number divisible by $36$ is also divisible by $6$,the number of $3$-digit numbers divisible by $2$ and $3$ but not by $4$ and $9$ is $150 - 25 = 125$.

(B) To find the largest exponent $n$ such that $3^n$ divides $50!$,we use Legendre's Formula: $E_p(m!) = \sum_{k=1}^{\infty} \left[ \frac{m}{p^k} \right]$.
Here,$m = 50$ and $p = 3$.
$n = \left[ \frac{50}{3} \right] + \left[ \frac{50}{3^2} \right] + \left[ \frac{50}{3^3} \right] + \left[ \frac{50}{3^4} \right]$
$n = \left[ \frac{50}{3} \right] + \left[ \frac{50}{9} \right] + \left[ \frac{50}{27} \right] + \left[ \frac{50}{81} \right]$
$n = 16 + 5 + 1 + 0$
$n = 22$.
Thus,the largest $n$ is $22$.

(C) We need to form a sequence of $10$ terms using the digits $0, 1, 2$ such that there are exactly five $1$s,three $2$s,and consequently two $0$s $(10 - 5 - 3 = 2)$.
This is a problem of permutations of a multiset where the items are not all distinct.
The number of ways to arrange $5$ ones,$3$ twos,and $2$ zeros is given by the multinomial coefficient:
$\text{Number of sequences} = \frac{10!}{5! \times 3! \times 2!}$
Calculating the value:
$\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times (3 \times 2 \times 1) \times (2 \times 1)} = \frac{10 \times 9 \times 8 \times 7 \times 6}{6 \times 2} = 10 \times 9 \times 4 \times 7 = 2520$.

(B) Let the $3$-digit number be $n$. We are given that $\text{gcd}(n, 36) = 2$.
Since $36 = 2^2 \times 3^2$,the condition $\text{gcd}(n, 36) = 2$ implies that $n$ must be divisible by $2$ but not by $4$,and $n$ must not be divisible by $3$.
Let $n = 2k$. Then $\text{gcd}(2k, 36) = 2 \implies \text{gcd}(k, 18) = 1$.
This means $k$ is not divisible by $2$ and not divisible by $3$.
The $3$-digit numbers are in the range $[100, 999]$.
So,$100 \le 2k \le 999 \implies 50 \le k \le 499.5$.
Thus,$k \in \{50, 51, \dots, 499\}$.
The total number of values for $k$ is $499 - 50 + 1 = 450$.
We need to exclude values of $k$ that are divisible by $2$ or $3$.
Let $S = \{50, 51, \dots, 499\}$.
Number of multiples of $2$ in $S$: $\lfloor \frac{499}{2} \rfloor - \lfloor \frac{49}{2} \rfloor = 249 - 24 = 225$.
Number of multiples of $3$ in $S$: $\lfloor \frac{499}{3} \rfloor - \lfloor \frac{49}{3} \rfloor = 166 - 16 = 150$.
Number of multiples of $6$ in $S$: $\lfloor \frac{499}{6} \rfloor - \lfloor \frac{49}{6} \rfloor = 83 - 8 = 75$.
By the Principle of Inclusion-Exclusion,the number of $k$ divisible by $2$ or $3$ is $225 + 150 - 75 = 300$.
The number of $k$ such that $\text{gcd}(k, 6) = 1$ is $450 - 300 = 150$.

(D) Let $d$ be the smallest divisor of a composite number $a$ such that $1 < d < a$.
If $d > \sqrt{a}$,then the other divisor $a/d$ must also be greater than $\sqrt{a}$ because $a/d < a/\sqrt{a} = \sqrt{a}$.
This implies $a/d$ is a divisor smaller than $d$,which contradicts the assumption that $d$ is the smallest divisor greater than $1$.
Therefore,the smallest divisor $d$ must satisfy $d \leq \sqrt{a}$.

(C) We know that,if $a = p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot p_{3}^{\alpha_{3}} \dots$
Then,the total number of positive divisors of $a$ is given by $T(a) = (\alpha_{1} + 1)(\alpha_{2} + 1)(\alpha_{3} + 1) \dots$
Given,$252 = 2^{2} \times 3^{2} \times 7^{1}$
Here,$\alpha_{1} = 2, \alpha_{2} = 2, \alpha_{3} = 1$
$\therefore T(252) = (2 + 1)(2 + 1)(1 + 1)$
$= 3 \cdot 3 \cdot 2$
$= 18$

(B) To find the greatest common divisor $(24, 92)$,we use the Euclidean algorithm:
$92 = 3 \times 24 + 20$
$24 = 1 \times 20 + 4$
$20 = 5 \times 4 + 0$
Thus,$(24, 92) = 4$.
Now,express $4$ as a linear combination of $24$ and $92$:
$4 = 24 - 1 \times 20$
Substitute $20 = 92 - 3 \times 24$:
$4 = 24 - 1 \times (92 - 3 \times 24)$
$4 = 24 - 92 + 3 \times 24$
$4 = 4 \times 24 - 1 \times 92$
Comparing this with $24m + 92n = 4$,we get $m = 4$ and $n = -1$.
Therefore,$(m, n) = (4, -1)$.

(A) To find the last digit of $7^{886}$,we observe the pattern of the last digits of powers of $7$:
$7^{1} = 7$
$7^{2} = 49$ (last digit $9$)
$7^{3} = 343$ (last digit $3$)
$7^{4} = 2401$ (last digit $1$)
The cycle of last digits is $(7, 9, 3, 1)$ with a period of $4$.
We divide the exponent $886$ by $4$:
$886 = 4 \times 221 + 2$
Thus,$7^{886} = (7^{4})^{221} \times 7^{2}$.
The last digit of $(7^{4})^{221}$ is $1^{221} = 1$.
The last digit of $7^{2}$ is $9$.
Therefore,the last digit of $7^{886}$ is $1 \times 9 = 9$.

(A) The prime factorization of $242$ is $2 \times 11^2$.
The divisors of $242$ are $1, 2, 11, 22, 121, 242$.
The divisors excluding $1$ and $242$ are $2, 11, 22, 121$.
The sum of these divisors is $2 + 11 + 22 + 121 = 156$.

(D) We observe the powers of $5$:
$5^{1} = 5$
$5^{2} = 25$
$5^{3} = 125$
$5^{4} = 625$
It is evident that for any positive integer $n$,the unit digit of $5^{n}$ is always $5$.
Therefore,the unit's place of $5^{834}$ is $5$.

(A) The number of terms in the expansion of $(x_1 + x_2 + \dots + x_r)^n$ is given by the formula $\binom{n+r-1}{r-1}$.
Here,$n = 10$ and $r = 3$.
Substituting these values,we get:
$\binom{10+3-1}{3-1} = \binom{12}{2}$.
Calculating the value: $\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66$.

(C) The condition $a \equiv b \pmod{x}$ implies that $x$ is a divisor of $(a - b)$.
For the first congruence: $21 \equiv 385 \pmod{x}$,we have $x$ divides $(385 - 21) = 364$.
The divisors of $364$ are $1, 2, 4, 7, 13, 14, 26, 28, 52, 91, 182, 364$.
For the second congruence: $587 \equiv 167 \pmod{x}$,we have $x$ divides $(587 - 167) = 420$.
The divisors of $420$ are $1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, 420$.
The greatest common divisor of $364$ and $420$ is $gcd(364, 420) = 28$.
Thus,the greatest value of $x$ is $28$.