For natural numbers $n$ with $2 \leq n \leq 100$, find the number of values of $n$ such that $\left( \sqrt[3]{3^5} \right)^{\frac{1}{2}}$ is the $n$-th root of some natural number. [4 points]

For a natural number $m$ ($m \geq 2$), let $f ( m )$ be the number of natural numbers $n \geq 2$ such that an integer exists among the $n$-th roots of $m ^ { 12 }$. What is the value of $\sum _ { m = 2 } ^ { 9 } f ( m )$? [4 points]
(1) 37
(2) 42
(3) 47
(4) 52
(5) 57

$50^{\text{th}}$ root of a number $x$ is 12 and $50^{\text{th}}$ root of another number $y$ is 18. Then the remainder obtained on dividing $(x + y)$ by 25 is $\_\_\_\_$.

Let $\alpha = \frac { ( 4 ! ) ! } { ( 4 ! ) ^ { 3 ! } }$ and $\beta = \frac { ( 5 ! ) ! } { ( 5 ! ) ^ { 4 ! } }$. Then :
(1) $\alpha \in \mathrm { N }$ and $\beta \notin \mathrm { N }$
(2) $\alpha \notin \mathrm { N }$ and $\beta \in \mathrm { N }$
(3) $\alpha \in \mathrm { N }$ and $\beta \in \mathrm { N }$
(4) $\alpha \notin \mathrm { N }$ and $\beta \notin \mathrm { N }$

Where $m$ and $n$ are positive integers, consider the rational number
$$r = \frac { m } { 3 } + \frac { n } { 7 } .$$
We are to find $m$ and $n$ such that among all $r$'s satisfying $r < \sqrt { 2 }$, $r$ is closest to $\sqrt { 2 }$.
It is sufficient that among all $m$'s and $n$'s which satisfy the inequality
$$\mathbf { A } m + \mathbf { B } n < \mathbf { CD } \sqrt { 2 }$$
we find the $m$ and $n$ such that $\mathrm { A } m + \mathrm { B } n$ is closest to $\mathrm { CD } \sqrt { 2 }$.
Squaring both sides of (1), we have
$$(\mathrm { A } m + \mathrm { B } n ) ^ { 2 } < \mathbf { EFG } .$$
Here, the greatest square number which is smaller than EFG is $\mathbf{HIJ} = \mathbf{KL}^{2}$. So, let us consider the equation
$$\mathrm { A } m + \mathrm { B } n = \mathrm { KL } .$$
Transforming this equation, we have
$$n = \frac { \mathbf { MN } - \mathbf { O } m } { \mathbf { P } } .$$
Since $n$ is an integer, $\mathbf { MN } - \mathbf { O } m$ is a multiple of $\mathbf { Q }$. Thus, we obtain
$$m = \mathbf { R } , \quad n = \mathbf { S } .$$

Let $m$ and $n$ be positive integers satisfying $0 < m - n\sqrt{2} < 1$. Denote the integral part of $(m + n\sqrt{2})^3$ by $a$ and the fractional part by $b$.
(1) We are to prove that $a$ is an odd number and $(m - n\sqrt{2})^3 = 1 - b$.
If $(m + n\sqrt{2})^3 = p + q\sqrt{2}$, where $p$ and $q$ are integers, then $$p = m^3 + \mathbf{A}mn^2, \quad q = \mathbf{B}m^2n + \mathbf{C}n^3.$$ Thus, we see that $(m - n\sqrt{2})^3 = p - q\sqrt{2}$.
Furthermore, the integral part of $(m - n\sqrt{2})^3$ is $\mathbf{D}$. When we denote its fractional part by $c$, the following two equations hold: $$\left\{\begin{array}{l} p + q\sqrt{2} = a + b \\ p - q\sqrt{2} = c \end{array}\right.$$
From these we obtain $$\mathbf{E} \quad p - a = b + c.$$
Here, since the left side is an integer and the range of values which the right side takes is $\mathbf{F} < b + c < \mathbf{G}$, we see that $$b + c = \mathbf{H}$$
Hence we see that $a = \mathbf{E}p - \mathbf{H}$, which shows that $a$ is an odd number and that $(m - n\sqrt{2})^3 = 1 - b$.
(2) Let us find the values of $m$ and $n$ when $a = 197$.
Since $a = 197$, we see that $p = \mathbf{IJ}$, that is, $m^3 + \mathbf{A}mn^2 = \mathbf{IJ}$. The positive integers $m$ and $n$ satisfying this equation are $$m = \mathbf{K}, \quad n = \mathbf{L}.$$

16. Given that $c$ and $d$ are non-zero integers, the expression $\frac { 10 ^ { c - 2 d } \times 20 ^ { 2 c + d } } { 8 ^ { c } \times 125 ^ { c + d } }$ is an integer if
A $\quad c < 0$
B $\quad d < 0$
C $\quad c < 0$ and $d < 0$
D $\quad c < 0$ and $d > 0$
E $\quad c > 0$ and $d < 0$
F $\quad c > 0$ and $d > 0$ G $\quad d > 0$
H $\quad c > 0$

For a project, 16 schools were selected from each of Turkey's 81 provinces, and a message was sent to each school's principal. Then, each school's principal sent this message to 35 teachers in their school.
Accordingly, what is the total number of principals and teachers to whom this message was sent?
A) $4^{6}$
B) $5^{6}$
C) $6^{6}$
D) $7 \cdot 5^{5}$
E) $8^{6}$

For distinct natural numbers $a$, $b$, and $c$,
$$\frac { 6 ^ { a } \cdot 15 ^ { b } } { 9 ^ { b } \cdot 10 ^ { c } }$$
is equal to an integer. Accordingly, which of the following orderings is correct?
A) $a < b < c$
B) $b < a < c$
C) $b < c < a$
D) $c < a < b$
E) $c < b < a$

Let $A$ and $B$ be natural numbers. A square with side length $A\sqrt{B}$ units has an area of 720 square units.
Accordingly, which of the following cannot be the sum $A + B$?
A) 26 B) 49 C) 83 D) 127 E) 182