Divisibility problems Questions in English

(A) Given $X = \{ 8^n - 7n - 1 : n \in N \}$ and $Y = \{ 49(n - 1) : n \in N \}$.
Using the binomial expansion,$8^n = (1 + 7)^n = 1 + ^nC_1(7) + ^nC_2(7^2) + ^nC_3(7^3) + \dots + ^nC_n(7^n)$.
Therefore,$8^n - 7n - 1 = (1 + 7n + ^nC_2(7^2) + ^nC_3(7^3) + \dots + 7^n) - 7n - 1 = ^nC_2(7^2) + ^nC_3(7^3) + \dots + 7^n$.
This simplifies to $49 \times [^nC_2 + ^nC_3(7) + \dots + 7^{n-2}]$.
For $n=1$,$8^1 - 7(1) - 1 = 0$.
For $n=2$,$8^2 - 7(2) - 1 = 64 - 14 - 1 = 49$.
For $n=3$,$8^3 - 7(3) - 1 = 512 - 21 - 1 = 490 = 49 \times 10$.
Thus,every element in $X$ is a multiple of $49$ (including $0$).
$Y = \{ 0, 49, 98, 147, \dots \}$ represents all non-negative multiples of $49$.
Since every element of $X$ is a multiple of $49$,$X \subseteq Y$.

(B) Using the Binomial Theorem,we expand $(1 + x)^n$ as follows:
$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$
Now,subtract $nx + 1$ from the expansion:
$(1 + x)^n - nx - 1 = (1 + nx + \frac{n(n - 1)}{2}x^2 + \frac{n(n - 1)(n - 2)}{6}x^3 + \dots) - nx - 1$
Simplifying the expression:
$(1 + x)^n - nx - 1 = x^2 \left[ \frac{n(n - 1)}{2} + \frac{n(n - 1)(n - 2)}{6}x + \dots \right]$
Since the expression is a multiple of $x^2$,it is divisible by $x^2$.

(C) Using the Binomial Theorem,we expand $(1 + 100)^{100}$:
$(1 + 100)^{100} = 1 + \binom{100}{1}(100) + \binom{100}{2}(100)^2 + \binom{100}{3}(100)^3 + \dots$
$(1 + 100)^{100} = 1 + 100(100) + \frac{100 \times 99}{2}(100)^2 + \dots$
$(1 + 100)^{100} = 1 + 10000 + \frac{9900}{2}(10000) + \dots$
Subtracting $1$ from both sides:
$101^{100} - 1 = 10000 + 4950(10000) + \dots$
$101^{100} - 1 = 10000(1 + 4950 + \dots)$
Thus,the number $101^{100} - 1$ is divisible by $10000$.

(A) For any $n \in N$,the expression ${x^{2n - 1}} + {y^{2n - 1}}$ represents the sum of two terms with the same odd exponent.
According to the algebraic identity,${a^k} + {b^k}$ is divisible by $a + b$ if $k$ is an odd positive integer.
Here,$k = 2n - 1$,which is always an odd integer for any $n \in N$.
Therefore,${x^{2n - 1}} + {y^{2n - 1}}$ is always divisible by $x + y$.

(A) Let $f(n) = 7^{2n} + 2^{3n-3} \cdot 3^{n-1}$.
For $n = 1$,$f(1) = 7^{2(1)} + 2^{3(1)-3} \cdot 3^{1-1} = 7^2 + 2^0 \cdot 3^0 = 49 + 1 = 50$.
For $n = 2$,$f(2) = 7^{2(2)} + 2^{3(2)-3} \cdot 3^{2-1} = 7^4 + 2^3 \cdot 3^1 = 2401 + 8 \cdot 3 = 2401 + 24 = 2425$.
Since $50$ and $2425$ are both divisible by $25$,the expression is always divisible by $25$.

(C) Let $f(n) = 11^{n+2} + 12^{2n+1}$.
For $n = 1$,$f(1) = 11^{1+2} + 12^{2(1)+1} = 11^3 + 12^3 = 1331 + 1728 = 3059$.
Dividing $3059$ by $133$,we get $3059 \div 133 = 23$.
Since $3059$ is divisible by $133$,the expression is divisible by $133$ for $n=1$.
For $n=2$,$f(2) = 11^4 + 12^5 = 14641 + 248832 = 263473$.
Dividing $263473$ by $133$,we get $263473 \div 133 = 1981$.
Thus,the expression is divisible by $133$ for all $n \in N$.

(B) The expression is $n(n^2 - 1) = (n - 1)n(n + 1)$.
This is the product of three consecutive natural numbers.
The product of $k$ consecutive natural numbers is always divisible by $k!$.
Therefore,the product of $3$ consecutive natural numbers is divisible by $3! = 3 \times 2 \times 1 = 6$.

(C) Let $f(x) = x^n - nax^{n-1} + (n-1)a^n$.
To check divisibility by $(x-a)^2$,we check if $f(a) = 0$ and $f'(a) = 0$.
$f(a) = a^n - na(a^{n-1}) + (n-1)a^n = a^n - na^n + na^n - a^n = 0$.
$f'(x) = nx^{n-1} - n(n-1)ax^{n-2}$.
$f'(a) = na^{n-1} - n(n-1)a(a^{n-2}) = na^{n-1} - n(n-1)a^{n-1} = na^{n-1} - (n^2-n)a^{n-1} = (n - n^2 + n)a^{n-1} = (2n - n^2)a^{n-1}$.
For $f'(a) = 0$,we need $n(2-n) = 0$,so $n=2$.
However,checking the original expression $x^n - nax^{n-1} + (n-1)a^n$ for $n=1$: $x - a + 0 = x-a$ (not divisible by $(x-a)^2$).
For $n=2$: $x^2 - 2ax + a^2 = (x-a)^2$ (divisible).
For $n=3$: $x^3 - 3ax^2 + 2a^3 = (x-a)^2(x+2a)$ (divisible).
Thus,the expression is divisible by $(x-a)^2$ for all $n \in N$ where $n \ge 2$.

(B) We need to find the remainder of $5^{99}$ when divided by $13$.
Using Fermat's Little Theorem,since $13$ is a prime number and $\gcd(5, 13) = 1$,we have $5^{13-1} \equiv 1 \pmod{13}$,which means $5^{12} \equiv 1 \pmod{13}$.
We can write $5^{99} = 5^{12 \times 8 + 3} = (5^{12})^8 \times 5^3$.
Since $5^{12} \equiv 1 \pmod{13}$,then $(5^{12})^8 \equiv 1^8 \equiv 1 \pmod{13}$.
Thus,$5^{99} \equiv 1 \times 5^3 \pmod{13}$.
$5^3 = 125$.
Dividing $125$ by $13$: $125 = 13 \times 9 + 8$.
Therefore,$125 \equiv 8 \pmod{13}$.
The remainder is $8$.

(C) We know that $2^2 = 4 \equiv -1 \pmod{5}$.
Squaring both sides,we get $(2^2)^2 \equiv (-1)^2 \pmod{5}$,which implies $2^4 \equiv 1 \pmod{5}$.
We can write $2^{301}$ as $2^{300} \times 2 = (2^4)^{75} \times 2$.
Since $2^4 \equiv 1 \pmod{5}$,then $(2^4)^{75} \equiv 1^{75} \equiv 1 \pmod{5}$.
Therefore,$2^{301} \equiv 1 \times 2 \equiv 2 \pmod{5}$.
The least positive remainder is $2$.

(C) Let $P(n) = {10^n} + 3 \times {4^{n + 2}} + 5$.
For $n = 1$,$P(1) = {10^1} + 3 \times {4^3} + 5 = 10 + 3 \times 64 + 5 = 10 + 192 + 5 = 207$.
Since $207 = 9 \times 23$,it is divisible by $9$.
For $n = 2$,$P(2) = {10^2} + 3 \times {4^4} + 5 = 100 + 3 \times 256 + 5 = 100 + 768 + 5 = 873$.
Since $873 = 9 \times 97$,it is divisible by $9$.
Thus,the expression is divisible by $9$ for all $n \in N$.

(A) Let $P(n) = 3^{2n+2} - 8n - 9$.
For $n=1$,$P(1) = 3^4 - 8(1) - 9 = 81 - 17 = 64$,which is divisible by $16$.
For $n=2$,$P(2) = 3^6 - 8(2) - 9 = 729 - 16 - 9 = 704$.
$704 = 16 \times 44$,so it is divisible by $16$.
Using the Binomial Theorem: $3^{2n+2} = 9 \times 3^{2n} = 9 \times (1+8)^n$.
$9 \times (1+8)^n = 9 \times [1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots]$.
$9 \times [1 + 8n + 32n(n-1) + \dots] = 9 + 72n + 288n(n-1) + \dots$.
Thus,$P(n) = 9 + 72n + 288n(n-1) + \dots - 8n - 9 = 64n + 288n(n-1) + \dots$.
Since every term is divisible by $64$,the expression is divisible by $64$,and consequently by $16$.

(D) We need to find the remainder of $17^{30}$ when divided by $5$.
First,observe that $17 \equiv 2 \pmod{5}$.
Therefore,$17^{30} \equiv 2^{30} \pmod{5}$.
By Fermat's Little Theorem,since $5$ is a prime number and $\gcd(2, 5) = 1$,we have $2^{5-1} \equiv 1 \pmod{5}$,which means $2^4 \equiv 1 \pmod{5}$.
Now,we can write $2^{30} = (2^4)^7 \times 2^2$.
Substituting the congruence,$2^{30} \equiv (1)^7 \times 2^2 \pmod{5}$.
$2^{30} \equiv 1 \times 4 \pmod{5}$.
$2^{30} \equiv 4 \pmod{5}$.
Thus,the least remainder is $4$.

(A) We need to find the remainder when $8^{2n} - 62^{2n+1}$ is divided by $9$.
Note that $8 \equiv -1 \pmod{9}$ and $62 \equiv 8 \equiv -1 \pmod{9}$.
Substituting these into the expression:
$8^{2n} - 62^{2n+1} \equiv (-1)^{2n} - (-1)^{2n+1} \pmod{9}$.
Since $2n$ is an even integer,$(-1)^{2n} = 1$.
Since $2n+1$ is an odd integer,$(-1)^{2n+1} = -1$.
Therefore,$1 - (-1) = 1 + 1 = 2$.
Thus,the remainder is $2$.

(D) For Statement $-2$: By Fermat's Little Theorem,for any prime $p$ and integer $n$,$n^p \equiv n \pmod{p}$. Here $p = 7$,so $n^7 \equiv n \pmod{7}$,which means $n^7 - n$ is divisible by $7$. Thus,Statement $-2$ is true.
For Statement $-1$: Expand $(n + 1)^7$ using the Binomial Theorem: $(n + 1)^7 = n^7 + 7n^6 + 21n^5 + 35n^4 + 35n^3 + 21n^2 + 7n + 1$.
Then $(n + 1)^7 - n^7 - 1 = 7n^6 + 21n^5 + 35n^4 + 35n^3 + 21n^2 + 7n = 7(n^6 + 3n^5 + 5n^4 + 5n^3 + 3n^2 + n)$.
This expression is clearly divisible by $7$. Thus,Statement $-1$ is true.
Note that Statement $-1$ can be written as $(n+1)^7 - (n+1) - (n^7 - n) = 7k$. Since both $(n+1)^7 - (n+1)$ and $n^7 - n$ are divisible by $7$ (by Statement $-2$),their difference is also divisible by $7$. Therefore,Statement $-2$ is a correct explanation for Statement $-1$.

(A) According to the Remainder Theorem,when a polynomial $P(x)$ is divided by $(x - a)$,the remainder is $P(a)$.
Here,$P(x) = x^{64} + x^{27} + 1$ and the divisor is $(x + 1)$,which is $(x - (-1))$.
Therefore,the remainder is $P(-1) = (-1)^{64} + (-1)^{27} + 1$.
Since $64$ is an even number,$(-1)^{64} = 1$.
Since $27$ is an odd number,$(-1)^{27} = -1$.
Thus,the remainder is $1 + (-1) + 1 = 1 - 1 + 1 = 1$.

(C) We use the binomial expansion: ${49^n} = {(1 + 48)^n}$.
Using the binomial theorem,${49^n} = 1 + n(48) + \frac{n(n-1)}{2}(48^2) + \dots + 48^n$.
Substituting this into the expression:
${49^n} + 16n - 1 = (1 + 48n + \frac{n(n-1)}{2}(2304) + \dots) + 16n - 1$.
$= 48n + 16n + \frac{n(n-1)}{2}(2304) + \dots$
$= 64n + 1152n(n-1) + \dots$
Since $1152 = 64 \times 18$,every term in the expansion is a multiple of $64$.
Thus,${49^n} + 16n - 1$ is divisible by $64$.

(B) Let the integer be $x$.
The expression is $x - x^3 = x(1 - x^2) = x(1 - x)(1 + x) = (x - 1)(x)(x + 1)$.
This is the product of three consecutive integers.
The product of $n$ consecutive integers is always divisible by $n!$.
Therefore,the product of $3$ consecutive integers is divisible by $3! = 3 \times 2 \times 1 = 6$.

(D) We need to find the last digit of $(3^P + 2)$ where $P = 3^{4n}$ and $n \in N$.
The last digit of powers of $3$ follows a cycle of $4$: $3^1 = 3$,$3^2 = 9$,$3^3 = 27$ (last digit $7$),$3^4 = 81$ (last digit $1$).
Since $P = 3^{4n}$,$P$ is a multiple of $4$ (specifically,$P$ is a power of $3$ which is always odd,but $3^{4n}$ is a multiple of $81$,so $P$ is a multiple of $4$ is incorrect; rather,$3^{4n}$ is of the form $(3^4)^n = 81^n$).
Any number ending in $1$ raised to any power $n$ still ends in $1$. Thus,$3^{4n} = (3^4)^n = 81^n$ ends in $1$.
Therefore,$3^P = 3^{(3^{4n})}$. Since $3^{4n}$ is a multiple of $4$,let $3^{4n} = 4k$ for some integer $k$.
Then $3^P = 3^{4k} = (3^4)^k = 81^k$,which ends in $1$.
Finally,the last digit of $(3^P + 2)$ is $1 + 2 = 3$ is incorrect. Let us re-evaluate: $P = 3^{4n}$. $3^{4n}$ is a power of $3$ which is always of the form $4k+1$ or $4k+3$. Actually,$3^{4n} = (3^4)^n = 81^n$. Since $81^n$ ends in $1$,$3^{4n}$ is of the form $4k+1$.
Thus $3^P = 3^{4k+1} = 3^{4k} \times 3^1$. The last digit of $3^{4k}$ is $1$,so the last digit of $3^P$ is $1 \times 3 = 3$.
Therefore,the last digit of $(3^P + 2)$ is $3 + 2 = 5$.

(C) We need to find the remainder when $(15^{23} + 23^{23})$ is divided by $19$.
Note that $15 \equiv -4 \pmod{19}$ and $23 \equiv 4 \pmod{19}$.
Therefore,$15^{23} + 23^{23} \equiv (-4)^{23} + 4^{23} \pmod{19}$.
Since $23$ is an odd integer,$(-4)^{23} = -4^{23}$.
Thus,$15^{23} + 23^{23} \equiv -4^{23} + 4^{23} \equiv 0 \pmod{19}$.
Hence,the remainder is $0$.

(A) We know that for any odd positive integer $n$,$a^n + b^n$ is divisible by $(a + b)$.
Here,$n = 27$,which is an odd number.
Therefore,$(11)^{27} + (21)^{27}$ is divisible by $(11 + 21) = 32$.
Since $32$ is a multiple of $16$ (i.e.,$32 = 2 \times 16$),any number divisible by $32$ must also be divisible by $16$.
Thus,$(11)^{27} + (21)^{27}$ is divisible by $16$.
When a number is perfectly divisible by the divisor,the remainder is $0$.

(C) We need to find the remainder when $3^{2003}$ is divided by $28$.
First,observe that $3^3 = 27 = 28 - 1$.
We can write $3^{2003} = 3^2 \cdot 3^{2001} = 9 \cdot (3^3)^{667}$.
Substituting $3^3 = 28 - 1$,we get $9 \cdot (28 - 1)^{667}$.
Using the Binomial Theorem,$(28 - 1)^{667} = \sum_{k=0}^{667} \binom{667}{k} (28)^{667-k} (-1)^k$.
This expression is of the form $28n + (-1)^{667} = 28n - 1$ for some integer $n$.
So,$3^{2003} = 9(28n - 1) = 28(9n) - 9$.
Since the remainder must be positive,we write $-9$ as $28 - 37$ or simply adjust: $28(9n - 1) + 19$.
Thus,the remainder is $19$.

(A) We need to find the remainder when $(13)^{507}$ is divided by $9$.
$(13)^{507} = (9 + 4)^{507}$.
Using the Binomial Theorem,$(9 + 4)^{507} = \sum_{k=0}^{507} \binom{507}{k} 9^{507-k} 4^k$.
All terms except the last term (where $k=507$) contain a factor of $9$.
Thus,$(13)^{507} \equiv 4^{507} \pmod{9}$.
Now,$4^1 \equiv 4 \pmod{9}$,$4^2 = 16 \equiv 7 \pmod{9}$,$4^3 = 64 \equiv 1 \pmod{9}$.
Since $4^3 \equiv 1 \pmod{9}$,we have $4^{507} = (4^3)^{169} \equiv (1)^{169} \equiv 1 \pmod{9}$.
Therefore,the remainder is $1$.

(B) Let the three consecutive odd numbers be $(2k-1)$,$(2k+1)$,and $(2k+3)$.
The sum of their squares is $S = (2k-1)^2 + (2k+1)^2 + (2k+3)^2$.
Expanding these terms: $S = (4k^2 - 4k + 1) + (4k^2 + 4k + 1) + (4k^2 + 12k + 9) = 12k^2 + 12k + 11$.
Increasing this sum by $1$,we get $S + 1 = 12k^2 + 12k + 12 = 12(k^2 + k + 1)$.
Since $k^2 + k = k(k+1)$ is always even (as it is the product of two consecutive integers),let $k(k+1) = 2m$.
Then $S + 1 = 12(2m + 1) = 24m + 12$.
This expression is clearly divisible by $12$,but not necessarily by $24$ (since $12$ is not divisible by $24$).
Thus,the sum is divisible by $12$ but not by $24$.

(C) We need to find the remainder of $3^{37}$ when divided by $80$.
We can write $3^{37} = 3 \times (3^4)^9$.
Since $3^4 = 81$,we have $3^{37} = 3 \times (81)^9$.
We can express $81$ as $(80 + 1)$.
So,$3^{37} = 3 \times (80 + 1)^9$.
Using the Binomial Theorem,$(80 + 1)^9 = \binom{9}{0} 80^9 + \binom{9}{1} 80^8 + \dots + \binom{9}{8} 80 + 1$.
This can be written as $80k + 1$ for some integer $k$.
Therefore,$3^{37} = 3(80k + 1) = 240k + 3$.
When $240k + 3$ is divided by $80$,the remainder is $3$.

(D) Let $n = 100$. The expression is $\sqrt {\frac{10^{2n}-1}{9} - 2\left( \frac{10^n-1}{9} \right)}$.
$= \sqrt {\frac{10^{2n}-1 - 2 \times 10^n + 2}{9}} = \sqrt {\frac{10^{2n} - 2 \times 10^n + 1}{9}}$.
$= \sqrt {\left( \frac{10^n-1}{3} \right)^2} = \frac{10^n-1}{3}$.
Since $\frac{10^{100}-1}{9} = \underbrace{111...1}_{100\,\text{digits}}$,then $\frac{10^{100}-1}{3} = 3 \times \underbrace{111...1}_{100\,\text{digits}} = \underbrace{333...3}_{100\,\text{digits}}$.

(D) We need to find the remainder of $3^{91}$ when divided by $80$.
First,express $3^{91}$ in terms of powers related to $80$.
Note that $3^4 = 81 = 80 + 1$.
We can write $3^{91} = 3^3 \times 3^{88} = 27 \times (3^4)^{22}$.
Substituting $3^4 = 80 + 1$,we get $3^{91} = 27 \times (80 + 1)^{22}$.
Using the Binomial Theorem,$(80 + 1)^{22} = \sum_{k=0}^{22} \binom{22}{k} 80^k (1)^{22-k} = 1 + 22(80) + \dots + 80^{22}$.
Thus,$(80 + 1)^{22} = 80m + 1$ for some integer $m$.
So,$3^{91} = 27(80m + 1) = 27 \times 80m + 27$.
When this expression is divided by $80$,the remainder is $27$.

(D) We need to find the remainder when $(27)^{999}$ is divided by $7$.
We can write $27$ as $(28 - 1)$.
So,$(27)^{999} = (28 - 1)^{999}$.
Using the Binomial Theorem,$(x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$.
$(28 - 1)^{999} = \binom{999}{0} (28)^{999} (-1)^0 + \binom{999}{1} (28)^{998} (-1)^1 + \dots + \binom{999}{999} (28)^0 (-1)^{999}$.
Every term except the last one contains a factor of $28$,which is divisible by $7$.
$(27)^{999} = 7k + (-1)^{999}$,where $k$ is an integer.
$(27)^{999} = 7k - 1$.
To find the positive remainder,we write $7k - 1 = 7(k - 1) + 7 - 1 = 7(k - 1) + 6$.
Thus,the remainder is $6$.

(B) We need to find the remainder when $2^{403}$ is divided by $15$.
We can write $2^{403} = 2^3 \times 2^{400} = 8 \times (2^4)^{100} = 8 \times (16)^{100}$.
Since $16 = 15 + 1$,we have $2^{403} = 8(15 + 1)^{100}$.
Using the Binomial Theorem,$(15 + 1)^{100} = \sum_{r=0}^{100} \binom{100}{r} 15^r (1)^{100-r} = 1 + 100(15) + \binom{100}{2} 15^2 + \dots + 15^{100}$.
Thus,$2^{403} = 8[1 + 15(100 + \binom{100}{2} 15 + \dots)] = 8 + 15 \times [8(100 + \binom{100}{2} 15 + \dots)]$.
Therefore,$\frac{2^{403}}{15} = \frac{8}{15} + 8(100 + \binom{100}{2} 15 + \dots)$.
The fractional part is $\frac{8}{15}$,which is given as $\frac{k}{15}$.
Hence,$k = 8$.

For two numbers $a$ and $b$,if we can find numbers $q$ and $r$ such that $a = bq + r$,then we say that $b$ divides $a$ with $q$ as the quotient and $r$ as the remainder. Thus,to show that $6^{n} - 5n$ leaves a remainder of $1$ when divided by $25$,we prove that $6^{n} - 5n = 25k + 1$,where $k$ is some integer.
We have the binomial expansion:
$(1 + a)^{n} = {}^{n}C_{0} + {}^{n}C_{1}a + {}^{n}C_{2}a^{2} + \dots + {}^{n}C_{n}a^{n}$
For $a = 5$,we get:
$(1 + 5)^{n} = {}^{n}C_{0} + {}^{n}C_{1}(5) + {}^{n}C_{2}(5^{2}) + \dots + {}^{n}C_{n}(5^{n})$
$6^{n} = 1 + 5n + 25({}^{n}C_{2} + {}^{n}C_{3}(5) + \dots + 5^{n-2})$
$6^{n} - 5n = 1 + 25({}^{n}C_{2} + 5 \cdot {}^{n}C_{3} + \dots + 5^{n-2})$
Let $k = {}^{n}C_{2} + 5 \cdot {}^{n}C_{3} + \dots + 5^{n-2}$.
Then $6^{n} - 5n = 25k + 1$.
This shows that when $6^{n} - 5n$ is divided by $25$,the remainder is $1$.

To show that $9^{n+1}-8n-9$ is divisible by $64$,we need to prove that $9^{n+1}-8n-9 = 64k$,where $k$ is a natural number.
By the Binomial Theorem:
$(1+a)^{m} = \sum_{r=0}^{m} {^{m}C_{r}} a^{r} = {^{m}C_{0}} + {^{m}C_{1}}a + {^{m}C_{2}}a^{2} + \dots + {^{m}C_{m}}a^{m}$
For $a=8$ and $m=n+1$,we obtain:
$(1+8)^{n+1} = {^{n+1}C_{0}} + {^{n+1}C_{1}}(8) + {^{n+1}C_{2}}(8^{2}) + \dots + {^{n+1}C_{n+1}}(8^{n+1})$
$9^{n+1} = 1 + (n+1)(8) + 64 \left[ {^{n+1}C_{2}} + {^{n+1}C_{3}}(8) + \dots + {^{n+1}C_{n+1}}(8^{n-1}) \right]$
$9^{n+1} = 1 + 8n + 8 + 64 \left[ {^{n+1}C_{2}} + {^{n+1}C_{3}}(8) + \dots + {^{n+1}C_{n+1}}(8^{n-1}) \right]$
$9^{n+1} = 9 + 8n + 64 \left[ {^{n+1}C_{2}} + {^{n+1}C_{3}}(8) + \dots + {^{n+1}C_{n+1}}(8^{n-1}) \right]$
$9^{n+1} - 8n - 9 = 64k$,where $k = {^{n+1}C_{2}} + {^{n+1}C_{3}}(8) + \dots + {^{n+1}C_{n+1}}(8^{n-1})$ is a natural number.
Thus,$9^{n+1}-8n-9$ is divisible by $64$ for all positive integers $n$.

(C) We need to find the remainder when $(2021)^{3762}$ is divided by $17$.
First,note that $2021 = 17 \times 118 + 15$,or more conveniently,$2021 = 17 \times 119 - 2 = 2023 - 2$.
So,$(2021)^{3762} = (2023 - 2)^{3762}$.
Using the Binomial Theorem,$(2023 - 2)^{3762} = \sum_{k=0}^{3762} \binom{3762}{k} (2023)^{3762-k} (-2)^k$.
Since $2023 = 17 \times 119$,all terms containing $2023$ are divisible by $17$.
Thus,$(2021)^{3762} \equiv (-2)^{3762} \pmod{17}$.
$(-2)^{3762} = 2^{3762} = (2^4)^{940} \times 2^2 = 16^{940} \times 4$.
Since $16 \equiv -1 \pmod{17}$,we have $16^{940} \equiv (-1)^{940} \equiv 1 \pmod{17}$.
Therefore,$2^{3762} \equiv 1 \times 4 \equiv 4 \pmod{17}$.
The remainder is $4$.

(A) Given that the remainder when $x$ is divided by $4$ is $3$,we can write $x = 4k + 3$ for some integer $k$.
We need to find the remainder when $(2020 + x)^{2022}$ is divided by $8$.
Substitute $x = 4k + 3$ into the expression:
$(2020 + x)^{2022} = (2020 + 4k + 3)^{2022} = (2023 + 4k)^{2022}$.
Since $2023 = 8 \times 252 + 7$,we have $2023 \equiv 7 \equiv -1 \pmod{8}$.
Thus,$(2023 + 4k)^{2022} \equiv (-1 + 4k)^{2022} \pmod{8}$.
If $k$ is even,let $k = 2m$,then $(-1 + 4(2m))^{2022} = (-1 + 8m)^{2022} \equiv (-1)^{2022} \equiv 1 \pmod{8}$.
If $k$ is odd,let $k = 2m + 1$,then $(-1 + 4(2m + 1))^{2022} = (-1 + 8m + 4)^{2022} = (3 + 8m)^{2022} \equiv 3^{2022} \pmod{8}$.
Note that $3^2 = 9 \equiv 1 \pmod{8}$,so $3^{2022} = (3^2)^{1011} \equiv 1^{1011} \equiv 1 \pmod{8}$.
In both cases,the remainder is $1$.

(B) The given sum is a geometric progression with $a=1$,$r=3$,and $n=2022$ terms.
Sum $S = \frac{1(3^{2022}-1)}{3-1} = \frac{3^{2022}-1}{2}$.
We can write $3^{2022} = (3^2)^{1011} = 9^{1011} = (10-1)^{1011}$.
Using the binomial expansion,$(10-1)^{1011} = \binom{1011}{0} 10^{1011} - \binom{1011}{1} 10^{1010} + \ldots + \binom{1011}{1010} 10^1 (-1)^{1010} + (-1)^{1011}$.
$(10-1)^{1011} = 100k + 1011(10)(-1) - 1 = 100k - 10110 - 1 = 100k - 10111$.
Thus,$S = \frac{100k - 10111 - 1}{2} = \frac{100k - 10112}{2} = 50k - 5056$.
$S = 50k - 5050 - 6 = 50(k-101) - 6$.
Since the remainder must be positive,we write $S = 50(k-102) + 50 - 6 = 50(k-102) + 4$.
Therefore,the remainder is $4$.

(D) We need to find the remainder of $3^{2022}$ when divided by $5$.
$3^{2022} = (3^2)^{1011} = 9^{1011}$.
We can write $9$ as $(10 - 1)$.
So,$9^{1011} = (10 - 1)^{1011}$.
Using the Binomial Theorem,$(10 - 1)^{1011} = \sum_{k=0}^{1011} \binom{1011}{k} 10^{1011-k} (-1)^k$.
This can be written as $10 \cdot N + (-1)^{1011}$,where $N$ is an integer.
$10 \cdot N - 1 = 10 \cdot N - 5 + 4 = 5(2N - 1) + 4$.
Therefore,the remainder when $3^{2022}$ is divided by $5$ is $4$.

(C) We need to find the remainder of $(2021)^{2023}$ when divided by $7$.
First,divide $2021$ by $7$: $2021 = 7 \times 288 + 5$.
So,$2021 \equiv 5 \pmod{7}$,which is equivalent to $2021 \equiv -2 \pmod{7}$.
Therefore,$(2021)^{2023} \equiv (-2)^{2023} \pmod{7}$.
$(-2)^{2023} = - (2^{2023}) = - (2^3)^{674} \times 2^1$.
Since $2^3 = 8 \equiv 1 \pmod{7}$,we have $2^3 \equiv 1 \pmod{7}$.
Thus,$(2^3)^{674} \equiv 1^{674} \equiv 1 \pmod{7}$.
So,$(-2)^{2023} \equiv -(1 \times 2) \equiv -2 \pmod{7}$.
To express the remainder as a positive value,we add $7$: $-2 + 7 = 5$.
Thus,the remainder is $5$.

(D) We need to find the remainder when $(11)^{1011} + (1011)^{11}$ is divided by $9$.
First,observe that $11 \equiv 2 \pmod{9}$ and $1011 = 9 \times 112 + 3$,so $1011 \equiv 3 \pmod{9}$.
Thus,$(11)^{1011} + (1011)^{11} \equiv 2^{1011} + 3^{11} \pmod{9}$.
For $2^{1011} \pmod{9}$:
$2^6 = 64 \equiv 1 \pmod{9}$.
Since $1011 = 6 \times 168 + 3$,we have $2^{1011} = (2^6)^{168} \times 2^3 \equiv 1^{168} \times 8 \equiv 8 \pmod{9}$.
For $3^{11} \pmod{9}$:
$3^2 = 9 \equiv 0 \pmod{9}$,so $3^{11} = 3^2 \times 3^9 = 9 \times 3^9 \equiv 0 \pmod{9}$.
Therefore,$(11)^{1011} + (1011)^{11} \equiv 8 + 0 \equiv 8 \pmod{9}$.
The remainder is $8$.

(A) We need to find the remainder of $(2021)^{2022} + (2022)^{2021}$ when divided by $7$.
First,note that $2021 = 7 \times 288 + 5 \equiv -2 \pmod{7}$ and $2022 = 7 \times 288 + 6 \equiv -1 \pmod{7}$.
So,$(2021)^{2022} + (2022)^{2021} \equiv (-2)^{2022} + (-1)^{2021} \pmod{7}$.
Since $2022$ is even,$(-2)^{2022} = 2^{2022} = (2^3)^{674} = 8^{674} \equiv 1^{674} = 1 \pmod{7}$.
Since $2021$ is odd,$(-1)^{2021} = -1$.
Thus,$(2021)^{2022} + (2022)^{2021} \equiv 1 - 1 = 0 \pmod{7}$.
Therefore,the remainder is $0$.

(C) We need to find the remainder of $7^{2022} + 3^{2022}$ when divided by $5$.
Note that $7 \equiv 2 \pmod{5}$ and $3 \equiv -2 \pmod{5}$.
Thus,$7^{2022} + 3^{2022} \equiv 2^{2022} + (-2)^{2022} \pmod{5}$.
Since $2022$ is an even number,$(-2)^{2022} = 2^{2022}$.
So,$7^{2022} + 3^{2022} \equiv 2^{2022} + 2^{2022} = 2 \times 2^{2022} = 2^{2023} \pmod{5}$.
We know that $2^4 = 16 \equiv 1 \pmod{5}$.
Therefore,$2^{2023} = 2^{4 \times 505 + 3} = (2^4)^{505} \times 2^3 \equiv 1^{505} \times 8 \equiv 8 \pmod{5}$.
Finally,$8 \equiv 3 \pmod{5}$.
The remainder is $3$.

(D) Given $N_2 = 165 = 3 \times 5 \times 11$ and $N_1 = 2^{55} + 1$.
We know that if $n$ is an odd integer,then $x^n + y^n$ is divisible by $x + y$.
Since $55$ is odd,$N_1 = 2^{55} + 1^{55}$ is divisible by $2 + 1 = 3$.
Also,$N_1 = (2^5)^{11} + 1^{11} = 32^{11} + 1^{11}$,which is divisible by $32 + 1 = 33$.
Since $33 = 3 \times 11$,$N_1$ is divisible by both $3$ and $11$.
$N_2 = 165 = 15 \times 11 = 5 \times 33$.
Since $N_1$ is a multiple of $33$ and $N_2$ is a multiple of $33$,and $N_1$ is not divisible by $5$ (as $2^{55} + 1 \equiv (2^2)^{27} \times 2 + 1 \equiv (-1)^{27} \times 2 + 1 \equiv -2 + 1 \equiv -1 \equiv 4 \pmod{5}$),the $HCF$ of $N_1$ and $N_2$ is $33$.

(B) We analyze the remainder of $n$ when divided by $3$. Let $n \equiv r \pmod{3}$,where $r \in \{0, 1, 2\}$.
Case $1$: If $n \equiv 0 \pmod{3}$,then $n$ is a multiple of $3$. Consequently,$n^{19}$ is a multiple of $3$.
Case $2$: If $n \equiv 1 \pmod{3}$,then $n^{38} \equiv 1^{38} \equiv 1 \pmod{3}$. Thus,$n^{38} - 1 \equiv 0 \pmod{3}$,meaning $n^{38} - 1$ is a multiple of $3$.
Case $3$: If $n \equiv 2 \pmod{3}$,then $n^{38} \equiv 2^{38} \equiv (-1)^{38} \equiv 1 \pmod{3}$. Thus,$n^{38} - 1 \equiv 0 \pmod{3}$,meaning $n^{38} - 1$ is a multiple of $3$.
In all cases,at least one of the numbers in the set $\{n^{19}, n^{38}-1\}$ is a multiple of $3$. Therefore,the correct option is $B$.

(B) For any integer $n$,$n^3 \equiv 0, 1, \text{ or } 8 \pmod{9}$,which is equivalent to $n^3 \equiv 0, 1, -1 \pmod{9}$.
Statement $I$: We want to check if $a^3+b^3+c^3 \equiv 0 \pmod{9}$ implies $abc \equiv 0 \pmod{3}$.
If $abc$ is not divisible by $3$,then $a, b, c \not\equiv 0 \pmod{3}$. Thus $a, b, c \equiv 1, 2 \pmod{3}$.
Then $a^3, b^3, c^3 \equiv 1, 8 \pmod{9}$ (since $1^3=1$ and $2^3=8 \equiv -1 \pmod{9}$).
To have $a^3+b^3+c^3 \equiv 0 \pmod{9}$,we need the sum of three values from ${1, -1}$ to be $0 \pmod{9}$,which is impossible as the possible sums are $\{3, 1, -1, -3\}$.
Thus,at least one of $a, b, c$ must be a multiple of $3$,so $abc$ is divisible by $3$. Statement $I$ is true.
Statement $II$: Consider $a=1, b=2, c=1, d=2$.
Then $a^3+b^3+c^3+d^3 = 1^3+2^3+1^3+2^3 = 1+8+1+8 = 18$,which is divisible by $9$.
However,$abcd = 1 \times 2 \times 1 \times 2 = 4$,which is not divisible by $3$.
Thus,Statement $II$ is false.

(A) We need to find the remainder of $(2023)^{2023}$ when divided by $35$.
First,note that $2023 = 35 \times 57 + 28$,so $2023 \equiv 28 \equiv -7 \pmod{35}$.
Thus,$(2023)^{2023} \equiv (-7)^{2023} \pmod{35}$.
We can write $(-7)^{2023} = -7 \times 7^{2022} = -7 \times (7^2)^{1011} = -7 \times (49)^{1011}$.
Since $49 \equiv 14 \pmod{35}$,this does not simplify easily. Let's use $49 = 35 + 14$.
Alternatively,$7^{2022} = (7^2)^{1011} = 49^{1011} = (35 + 14)^{1011} \equiv 14^{1011} \pmod{35}$.
Since $14^2 = 196 = 35 \times 5 + 21$ and $14^3 = 14 \times 21 = 294 = 35 \times 8 + 14$,we see a cycle: $14^1 \equiv 14$,$14^2 \equiv 21$,$14^3 \equiv 14 \pmod{35}$.
For odd powers $n$,$14^n \equiv 14 \pmod{35}$.
Thus,$7^{2022} = (7^2)^{1011} = 49^{1011} \equiv 14^{1011} \equiv 14 \pmod{35}$.
Therefore,$(2023)^{2023} \equiv -7 \times 14 = -98 \pmod{35}$.
Since $-98 = -3 \times 35 + 7$,we have $-98 \equiv 7 \pmod{35}$.
The remainder is $7$.

(B) Given $x^{1/50} = 12 \implies x = 12^{50}$ and $y^{1/50} = 18 \implies y = 18^{50}$.
We need to find the remainder of $(12^{50} + 18^{50})$ divided by $25$.
$12^{50} = (12^2)^{25} = 144^{25} = (150 - 6)^{25}$.
Using the Binomial Theorem,$(150 - 6)^{25} = \sum_{k=0}^{25} \binom{25}{k} (150)^k (-6)^{25-k} = 25M + (-6)^{25}$.
$18^{50} = (18^2)^{25} = 324^{25} = (325 - 1)^{25}$.
Using the Binomial Theorem,$(325 - 1)^{25} = \sum_{k=0}^{25} \binom{25}{k} (325)^k (-1)^{25-k} = 25N + (-1)^{25} = 25N - 1$.
So,$x + y = 25M + (-6)^{25} + 25N - 1 = 25(M+N) - (6^{25} + 1)$.
Now,$6^{25} = (6^2)^{12} \cdot 6 = 36^{12} \cdot 6 = (25 + 11)^{12} \cdot 6$.
$(25 + 11)^{12} = 25P + 11^{12} = 25P + (121)^6 = 25P + (125 - 4)^6 = 25Q + (-4)^6 = 25Q + 4096$.
$4096 = 25 \times 163 + 21$.
Thus,$6^{25} = (25K + 21) \times 6 = 150K + 126 = 25(6K + 5) + 1$.
Therefore,$x + y = 25(M+N) - (25(6K+5) + 1 + 1) = 25R - 2$.
Since the remainder must be positive,we add $25$ to $-2$,giving $25 - 2 = 23$.

(A) By Fermat's Little Theorem,since $11$ is a prime number and $\gcd(5, 11) = 1$,we have $5^{10} \equiv 1 \pmod{11}$.
$5^{99} = 5^{90} \cdot 5^9 = (5^{10})^9 \cdot 5^9$.
Since $5^{10} \equiv 1 \pmod{11}$,then $(5^{10})^9 \equiv 1^9 \equiv 1 \pmod{11}$.
Now,we calculate $5^9 \pmod{11}$:
$5^2 = 25 \equiv 3 \pmod{11}$.
$5^4 = (5^2)^2 \equiv 3^2 = 9 \equiv -2 \pmod{11}$.
$5^8 = (5^4)^2 \equiv (-2)^2 = 4 \pmod{11}$.
$5^9 = 5^8 \cdot 5^1 \equiv 4 \cdot 5 = 20 \equiv 9 \pmod{11}$.
Therefore,$5^{99} \equiv 1 \cdot 9 = 9 \pmod{11}$.
The remainder is $9$.

(C) For $(S1):$ Let $a = 2023$ and $b = 1999$. Note that $a \equiv 7 \pmod{8}$ and $b \equiv 7 \pmod{8}$.
Then $a^{2022} \equiv 7^{2022} \pmod{8}$ and $b^{2022} \equiv 7^{2022} \pmod{8}$.
Since $7 \equiv -1 \pmod{8}$,$7^{2022} \equiv (-1)^{2022} \equiv 1 \pmod{8}$.
Thus,$2023^{2022} - 1999^{2022} \equiv 1 - 1 \equiv 0 \pmod{8}$. So,$(S1)$ is correct.
For $(S2):$ Let $f(n) = 13(13^{n}) - 11n - 13 = 13^{n+1} - 11n - 13$.
Using binomial expansion,$13^{n+1} = (1 + 12)^{n+1} = 1 + (n+1)12 + \binom{n+1}{2}12^2 + \dots = 1 + 12n + 12 + 144 \binom{n+1}{2} + \dots$
So,$f(n) = 1 + 12n + 12 + 144 \binom{n+1}{2} + \dots - 11n - 13 = n + 144 \binom{n+1}{2} + \dots$
For $f(n)$ to be divisible by $144$,we need $n$ to be a multiple of $144$. Since there are infinitely many such $n \in N$,$(S2)$ is correct.

(A) Let $E = 25^{190} - 19^{190} - 8^{190} + 2^{190}$.
We can rewrite this as $E = (25^{190} - 8^{190}) - (19^{190} - 2^{190})$.
Since $a^n - b^n$ is divisible by $a - b$ for any even $n$,we have:
$25^{190} - 8^{190}$ is divisible by $25 - 8 = 17$.
$19^{190} - 2^{190}$ is divisible by $19 - 2 = 17$.
Thus,$E$ is divisible by $17$.
Also,$E = (25^{190} - 19^{190}) - (8^{190} - 2^{190})$.
$25^{190} - 19^{190}$ is divisible by $25 - 19 = 6$.
$8^{190} - 2^{190}$ is divisible by $8 - 2 = 6$.
Thus,$E$ is divisible by $6$.
Since $E$ is divisible by $17$ and $6$,and $\text{gcd}(17, 6) = 1$,$E$ is divisible by $17 \times 2 = 34$.
Checking for $14$: $25 \equiv 4 \pmod{7}$,$19 \equiv 5 \pmod{7}$,$8 \equiv 1 \pmod{7}$,$2 \equiv 2 \pmod{7}$.
$E \equiv 4^{190} - 5^{190} - 1^{190} + 2^{190} \pmod{7}$.
Using Fermat's Little Theorem,$a^6 \equiv 1 \pmod{7}$. $190 = 6 \times 31 + 4$.
$E \equiv 4^4 - 5^4 - 1 + 2^4 \equiv 256 - 625 - 1 + 16 \equiv 4 - 2 - 1 + 2 \equiv 3 \pmod{7}$.
Since $E \not\equiv 0 \pmod{7}$,it is not divisible by $14$.
Therefore,it is divisible by $34$ but not by $14$.

(B) Let $N = (22)^{2022} + (2022)^{22}$.
For division by $3$:
$22 \equiv 1 \pmod{3}$,so $(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod{3}$.
$2022$ is divisible by $3$,so $(2022)^{22} \equiv 0^{22} \equiv 0 \pmod{3}$.
Thus,$N \equiv 1 + 0 \equiv 1 \pmod{3}$,which means $\alpha = 1$.
For division by $7$:
$22 \equiv 1 \pmod{7}$,so $(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod{7}$.
$2022 = 7 \times 288 + 6$,so $2022 \equiv 6 \equiv -1 \pmod{7}$.
$(2022)^{22} \equiv (-1)^{22} \equiv 1 \pmod{7}$.
Thus,$N \equiv 1 + 1 \equiv 2 \pmod{7}$,which means $\beta = 2$.
Therefore,$\alpha^2 + \beta^2 = 1^2 + 2^2 = 1 + 4 = 5$.

(B) We need to find the fractional part of $\frac{4^{2022}}{15}$.
Note that $4^{2022} = (4^2)^{1011} = 16^{1011}$.
We can write $16$ as $(15 + 1)$.
So,$16^{1011} = (15 + 1)^{1011}$.
Using the Binomial Theorem,$(15 + 1)^{1011} = \binom{1011}{0} 15^{1011} + \binom{1011}{1} 15^{1010} + \dots + \binom{1011}{1010} 15^1 + \binom{1011}{1011} 1^0$.
$(15 + 1)^{1011} = 15 \times K + 1$,where $K$ is an integer.
Therefore,$\frac{4^{2022}}{15} = \frac{15K + 1}{15} = K + \frac{1}{15}$.
The fractional part is $\left\{ \frac{4^{2022}}{15} \right\} = \frac{1}{15}$.

(B) We need to find the remainder of $7^{103} \pmod{17}$.
By Fermat's Little Theorem,$a^{p-1} \equiv 1 \pmod{p}$ if $p$ is prime and $p \nmid a$.
Here,$p=17$ and $a=7$,so $7^{16} \equiv 1 \pmod{17}$.
Now,$103 = 16 \times 6 + 7$.
Therefore,$7^{103} = (7^{16})^6 \times 7^7 \equiv 1^6 \times 7^7 \pmod{17}$.
$7^2 = 49 \equiv 15 \equiv -2 \pmod{17}$.
$7^4 \equiv (-2)^2 = 4 \pmod{17}$.
$7^6 \equiv 4 \times (-2) = -8 \equiv 9 \pmod{17}$.
$7^7 = 7^6 \times 7 \equiv 9 \times 7 = 63 \pmod{17}$.
Since $63 = 17 \times 3 + 12$,we have $63 \equiv 12 \pmod{17}$.
Thus,the remainder is $12$.