Let $n$ be a positive integer having 27 divisors including 1 and $n$, which are denoted by $d _ { 1 } , \ldots , d _ { 27 }$. Then the product of $d _ { 1 } , d _ { 2 } , \ldots , d _ { 27 }$ equals
(A) $n ^ { 13 }$.
(B) $n ^ { 14 }$.
(C) $n ^ { \frac { 27 } { 2 } }$.
(D) $27 n$.

The greatest positive integer $k$, for which $49 ^ { k } + 1$ is a factor of the sum $49 ^ { 125 } + 49 ^ { 124 } + \ldots + 49 ^ { 2 } + 49 + 1$, is
(1) 32
(2) 63
(3) 60
(4) 35

A natural number has prime factorization given by $n = 2 ^ { x } 3 ^ { y } 5 ^ { z }$, where $y$ and $z$ are such that $y + z = 5$ and $y ^ { - 1 } + z ^ { - 1 } = \frac { 5 } { 6 } , y > z$. Then the number of odd divisors of $n$, including 1 , is:
(1) 12
(2) 6
(3) 11
(4) $6 x$

The largest natural number $n$ such that $3 n$ divides 66! is $\_\_\_\_$

We are to find the natural number $a$ such that $3a + 1$ is a divisor of $a^2 + 5$.
Set $b = 3a + 1$. Then we have $$a^2 + 5 = \frac{b^2 - \mathbf{OO}b + \mathbf{PQ}}{\mathbf{R}}.$$
On the other hand, since $b$ is a divisor of $a^2 + 5$, $a^2 + 5$ can be expressed as $$a^2 + 5 = bc$$ for some natural number $c$. From (1) and (2), we have $$b(\mathbf{S}c - b + \mathbf{IT}) = \mathbf{UV}.$$
This shows that $b$ must also be one of the divisors of UV. Of these, only $b = \mathbf{WX}$ is a number such that $a$ is a natural number. Hence, $a = \mathbf{YZ}$.

Answer the following questions.
(1) The prime factorization of 1400 is $\mathbf{A}^{\mathbf{B}} \cdot \mathbf{C}^{\mathbf{D}} \cdot \mathbf{E}$ (give the answers in the order A/C).
(2) The number of the divisors of 1400 is $\mathbf{FG}$.
(3) Let $a$ and $b$ be any two divisors of 1400 satisfying $1 < a < b$. There are $\mathbf { H }$ pairs $( a , b )$ such that $a$ and $b$ are relatively prime and $a b = 1400$. Among them, $a$ and $b$ such that $b - a$ is maximized are
$$a = \mathbf { I } , \quad b = \mathbf { J K L } .$$
(4) For $a = \square$ and $b = \mathbf{JKL}$, consider the equation
We can transform (1) into the following equation:
$$y = \mathbf { M N } x + \frac { \mathbf { O } } { \mathbf { Q } } x - \frac { \mathbf { P } } { \mathbf { Q } } .$$
Therefore, among the pairs of positive integers $x$ and $y$ that satisfy equation (1), the pair such that $x$ is minimized is
$$x = \mathbf { R } , \quad y = \mathbf { S T } .$$

$$\prod _ { n = 1 } ^ { 7 } ( 3 n + 2 )$$
If this number is divisible by $10 ^ { \mathbf { m } }$, what is the maximum integer value that m can take?
A) 2
B) 3
C) 4
D) 5
E) 6

If the product $\mathrm{x} \cdot (10!)$ is the square of a positive integer, what is the smallest value that x can take?
A) 21 B) 7 C) 5 D) 10 E) 14

For $1 < n < 50$, how many integers n are there such that the number of positive divisors is 3?
A) 2
B) 3
C) 4
D) 5
E) 7

Let n be a positive integer. If every prime number p that divides n also divides $p ^ { 2 }$ into n, then n is called a powerful number.
Which of the following is NOT a powerful number?
A) 27
B) 64
C) 72
D) 99
E) 108