Divisibility problems Questions in English

(B) Let $x = 7^{7^{...7}}$ ($22$ times $7$). We want to find $x \pmod{48}$.
Note that $7^2 = 49 \equiv 1 \pmod{48}$.
Since the exponent of $7$ is $7^{7^{...7}}$ ($21$ times $7$),which is clearly an odd number,let the exponent be $2k+1$ for some integer $k$.
Then $7^{2k+1} = 7 \times (7^2)^k = 7 \times (49)^k$.
Since $49 \equiv 1 \pmod{48}$,we have $49^k \equiv 1^k \equiv 1 \pmod{48}$.
Therefore,$7 \times (49)^k \equiv 7 \times 1 \equiv 7 \pmod{48}$.
The remainder is $7$.

(A) Using the Binomial Theorem,we can write $(101)^{100}-1$ as $(1+100)^{100}-1$.
Expanding this using the binomial expansion:
$(1+100)^{100}-1 = \left(1 + {}^{100}C_{1}(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}\right) - 1$.
Since ${}^{100}C_{1} = 100$,the expression becomes:
$100(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}$.
$= 10^{4} + {}^{100}C_{2}(10^{4}) + {}^{100}C_{3}(10^{6}) + \dots + 10^{200}$.
$= 10^{4} \left(1 + {}^{100}C_{2} + {}^{100}C_{3}(10^{2}) + \dots + 10^{196}\right)$.
Thus,the expression is divisible by $10^{4}$.

(C) Using the Binomial Theorem,we have:
$(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + {^nC_3}x^3 + \dots + {^nC_n}x^n$
Since ${^nC_0} = 1$ and ${^nC_1} = n$,we can write:
$(1+x)^n = 1 + nx + {^nC_2}x^2 + {^nC_3}x^3 + \dots + x^n$
Now,subtract $nx$ and $1$ from both sides:
$(1+x)^n - nx - 1 = {^nC_2}x^2 + {^nC_3}x^3 + \dots + x^n$
Factoring out $x^2$ from the right side:
$(1+x)^n - nx - 1 = x^2 ({^nC_2} + {^nC_3}x + \dots + x^{n-2})$
Thus,the expression is divisible by $x^2$.

(C) We have $2^{3n} = (2^3)^n = 8^n$.
Since $8 = (1 + 7)$,we can write $8^n = (1 + 7)^n$.
Using the Binomial Theorem,$(1 + 7)^n = {^nC_0} + {^nC_1}(7) + {^nC_2}(7^2) + \dots + {^nC_n}(7^n)$.
Since ${^nC_0} = 1$,we have $8^n = 1 + 7({^nC_1} + {^nC_2}(7) + \dots + {^nC_n}(7^{n-1}))$.
Subtracting $1$ from both sides,$8^n - 1 = 7({^nC_1} + {^nC_2}(7) + \dots + {^nC_n}(7^{n-1}))$.
This expression is clearly divisible by $7$ for all $n \in N$.

(A) We know that $2^{3n} = (2^3)^n = 8^n$.
Since $8^n = (1+7)^n$,we can expand this using the Binomial Theorem:
$(1+7)^n = 1 + {}^{n}C_1(7) + {}^{n}C_2(7^2) + \dots + {}^{n}C_n(7^n)$.
Therefore,$2^{3n} - 1 = (1 + {}^{n}C_1(7) + {}^{n}C_2(7^2) + \dots + {}^{n}C_n(7^n)) - 1$.
$2^{3n} - 1 = 7({}^{n}C_1 + {}^{n}C_2(7) + \dots + {}^{n}C_n(7^{n-1}))$.
This expression is clearly divisible by $7$.

(B) We have the expression $2^{4n} - 15n - 1$.
Since $2^4 = 16$,we can write $2^{4n} = (16)^n = (1 + 15)^n$.
Using the Binomial Theorem,$(1 + 15)^n = {}^{n}C_{0} + {}^{n}C_{1}(15) + {}^{n}C_{2}(15)^2 + \dots + {}^{n}C_{n}(15)^n$.
Substituting this into the expression:
$2^{4n} - 15n - 1 = (1 + 15n + {}^{n}C_{2} \cdot 15^2 + \dots + {}^{n}C_{n} \cdot 15^n) - 15n - 1$.
Simplifying the terms,the $1$ and $15n$ cancel out:
$2^{4n} - 15n - 1 = {}^{n}C_{2} \cdot 15^2 + {}^{n}C_{3} \cdot 15^3 + \dots + {}^{n}C_{n} \cdot 15^n$.
Factoring out $15^2 = 225$:
$2^{4n} - 15n - 1 = 225 \cdot ({}^{n}C_{2} + {}^{n}C_{3} \cdot 15 + \dots + {}^{n}C_{n} \cdot 15^{n-2})$.
Thus,the expression is divisible by $225$ for all $n \in N$ where $n \ge 2$. For $n=1$,the expression is $2^4 - 15(1) - 1 = 16 - 15 - 1 = 0$,which is divisible by $225$.

(B) Statement $I$: We have $25^{13} + 3^{13} + 20^{13} + 8^{13}$.
Since $a^n + b^n$ is divisible by $(a + b)$ when $n$ is odd,we have:
$25^{13} + 3^{13}$ is divisible by $(25 + 3) = 28$,which is divisible by $7$.
$20^{13} + 8^{13}$ is divisible by $(20 + 8) = 28$,which is divisible by $7$.
Thus,the sum is divisible by $7$. Statement $I$ is true.
Statement $II$: Let $R = (7 + 4\sqrt{3})^{25} = I + f$,where $I$ is the integral part and $0 < f < 1$.
Let $R' = (7 - 4\sqrt{3})^{25} = f'$,where $0 < f' < 1$.
Since $7^2 - (4\sqrt{3})^2 = 49 - 48 = 1$,$R'$ is a very small positive number.
$R + R' = (7 + 4\sqrt{3})^{25} + (7 - 4\sqrt{3})^{25} = 2 \left[ {}^{25}C_0 7^{25} + {}^{25}C_2 7^{23}(4\sqrt{3})^2 + \dots \right]$.
This is an even integer. So,$I + f + f' = \text{even integer}$.
Since $0 < f + f' < 2$,$f + f'$ must be $1$.
Therefore,$I + 1 = \text{even integer}$,which implies $I$ is an odd integer. Statement $II$ is true.