Let $A = \{ 1,2,3 , \ldots , 10 \}$ and $B = \left\{ \frac { m } { n } : m , n \in A , m < n \right.$ and $\left. \operatorname { gcd } ( m , n ) = 1 \right\}$. Then $n ( B )$ is equal to:
(1) 36
(2) 31
(3) 37
(4) 29

Let $S = \{p_1, p_2, \ldots, p_{10}\}$ be the set of first ten prime numbers. Let $A = S \cup P$, where $P$ is the set of all possible products of distinct elements of $S$. Then the number of all ordered pairs $(x, y)$, $x \in S$, $y \in A$, such that $x$ divides $y$, is $\underline{\hspace{2cm}}$.

The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is $\underline{\hspace{2cm}}$.

Q2 Let $p$ be a prime number, and let $x$ and $y$ be positive integers. Then we are to find all triples of $p$, $x$ and $y$ which satisfy
$$\frac{p}{x} + \frac{7}{y} = p.$$
We can transform this equation into
$$(x - \mathbf{N})(py - \mathbf{OO}) = \mathbf{OP}.$$
From this, we obtain
$$x - \mathbf{N} = \mathbf{Q} \text{ or } \mathbf{R}, \quad (\text{note: have } \mathbf{Q} < \mathbf{R})$$
and hence
$$x = \mathbf{S} \text{ or } \mathbf{T}. \quad (\text{note: have } S < T)$$
First, if $x = S$, then
$$p = \mathbf{U}, \quad y = \mathbf{V}$$
or
$$p = \mathbf{W}, \quad y = \mathbf{X}. \quad (\text{note: have } \mathbf{U} < \mathbf{W})$$
Next, if $x = T$, then
$$p = \mathbf{Y}, \quad y = \mathbf{Z}.$$

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, $a = \mathbf{YZ}$.

Let $a$ and $b$ be natural numbers such that the greatest common divisor of $a$ and $b$ is 3. We are to find the natural numbers $a$ and $b$ such that
$$3 a - 2 b = \ell + 3$$
is satisfied, where $\ell$ is the least common multiple of $a$ and $b$.
When we set $a = 3 p$ and $b = 3 q$, the natural numbers $p$ and $q$ are mutually prime (co-prime), and hence $\ell = \mathbf { N } p q$. Thus using $p$ and $q$, the equality (1) can be transformed to
$$p q - \mathbf { O } p + \mathbf { P } q + \mathbf { Q } = 0 .$$
This can be further transformed to
$$( p + \mathbf { R } ) ( q - \mathbf { S } ) = - \mathbf { S } \mathbf { T } .$$
Among the pairs of integers $p$ and $q$ which satisfy this equation, the pair such that both $a$ and $b$ are natural numbers is
$$p = \mathbf { U } , \quad q = \mathbf { V } ,$$
which gives
$$a = \mathbf { W X } , \quad b = \mathbf { Y } .$$

We are to find a two-digit natural number $a$ such that $a + 9$ is a multiple of 7 and $a + 8$ is a multiple of 13.
First of all, $a + 9$ and $a + 8$ can be represented as
$$a + 9 = \mathbf { M } m , \quad a + 8 = \mathbf { N O } n ,$$
where $m$ and $n$ are natural numbers. From these two equalities, we have
$$\mathbf { M } m - \mathbf { N O } n = \mathbf { P } .$$
Since the pair of $m = \mathbf { Q }$ and $n = \mathbf { R }$ is an integral solution of (1), we have
$$\mathbf { M } ( m - \mathbf { Q } ) = \mathbf { NO } ( n - \mathbf { R } ) .$$
From (2), a natural number $n$ satisfying (1) can be represented as
$$n = \mathbf { S }$$
where $k$ is an integer. Thus
$$a = \mathbf { U V } k + \mathbf { W } ,$$
and since $a$ is a two-digit natural number, $a = \mathbf { X Y }$.

(1) Answer the following questions.
(i) Consider an integer $a$. When $a$ is divided by 5, the remainder is 4. So, $a$ can be represented as
$$a = \mathbf { A } \, k + \mathbf { B } \quad ( k \text{ : an integer}).$$
Hence, when $a ^ { 2 }$ is divided by 5, the remainder is $\mathbf { C }$.
(ii) The number written as the three-digit number $120_{(3)}$ in the base-3 system is $\mathbf{DE}$ in the decimal system.
The greatest natural number that can be expressed in three digits using the base-3 system is $\mathbf { F G }$ in the decimal system, and the smallest is $\mathbf { H }$ in the decimal system.
(2) For each of $\mathbf { I }$, $\mathbf { J }$ in the following statements, choose the correct answer from among (0) $\sim$ (3) below.
In the following, let $a$ be an integer and $b$ be a natural number.
(i) "When $a$ is divided by 5, the remainder is 4" is $\mathbf { I }$ for "when $a ^ { 2 }$ is divided by 5, the remainder is $\mathbf { C }$".
(ii) "$b$ satisfies $6 \leqq b \leqq 30$" is $\mathbf { J }$ for "$b$ is a three-digit number in the base-3 system".
(0) a necessary condition but not a sufficient condition
(1) a sufficient condition but not a necessary condition
(2) a necessary and sufficient condition
(3) neither a necessary condition nor a sufficient condition

Let $n$ be a positive integer, and $x$ and $y$ be non-negative integers. We are to examine the solutions of the following equation in $x$ and $y$
$$x ^ { 2 } - y ^ { 2 } = n . \tag{1}$$
First of all, by transforming (1), we obtain
$$( x + y ) ( x - y ) = n . \tag{2}$$
(1) When we find the solutions $( x , y )$ of (1) in the cases where $n = 8$ and $n = 9$, we have that if $n = 8$, then $( x , y ) = ( \mathbf { A } , \mathbf { B } )$, and if $n = 9$, then $( x , y ) = ( \mathbf { C } , \mathbf { D } ) , ( \mathbf { E } , \mathbf { F } )$. Note that you should write the solutions in the order such that $\mathbf{C} \leq \mathbf{E}$.
(2) For each of $\mathbf { G }$ $\sim$ $\mathbf { R }$ in the following sentences, choose the correct answer from among (0) $\sim$ (9) given below.
The following is a proof that (3) given below is the necessary and sufficient condition for (1) to have a solution.
Proof: First, suppose that $( x , y )$ satisfies (1). If $x$ and $y$ are both even or both odd, then both $x + y$ and $x - y$ are $\mathbf { G }$. Hence, by (2) we see that $n$ is a multiple of $\mathbf { H }$.
Next, if one of $x$ and $y$ is even and the other is odd, then both $x + y$ and $x - y$ are $\mathbf{I}$, and hence $n$ is $\mathbf{J}$.
Thus we see that
$$\text{``} n \text{ is a multiple of } \mathbf{H} \text{, or } n \text{ is } \mathbf{J} \text{''} \quad \ldots\ldots (3)$$
is a necessary condition for (1) to have a solution.
Conversely, suppose that $n$ satisfies the condition (3). If $n$ is a multiple of $\mathbf{H}$, then $n$ can be represented as $n = \mathbf{H} \cdot k$, where $k$ is a positive integer. So, if for example we take $x + y = \mathbf { K } \cdot k$ and $x - y = 2$, then $( x , y ) = ( k + \mathbf { L } , k - \mathbf { M } )$, which shows that (1) has a solution.
On the other hand, if $n$ is $\mathbf{J}$, then $n$ can be represented as $n = \mathbf { N } \ell + \mathbf { O }$, where $\ell$ is a non-negative integer. So, if for example we take $x + y = \mathbf { P } \ell + \mathbf { Q }$ and $x - y = 1$, then $( x , y ) = ( \ell + \mathbf { R } , \ell )$, which shows that (1) has a solution.
From the above, we see that the necessary and sufficient condition for (1) to have a solution is (3).
(0) 0 (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) even (8) odd (9) prime

Let $n$ be a two-digit natural number such that the remainder of $n ^ { 3 }$ divided by 66 is $n$. We are to find the number of such $n$ 's and to find the prime numbers among them.
From the conditions we have
$$n ^ { 3 } = \mathbf { A B } p + n \quad ( 0 < n \leqq \mathbf { C D } ) ,$$
where $p$ is the integer quotient of $n ^ { 3 }$ divided by 66 . This can be transformed into
$$n ( n - 1 ) ( n + 1 ) = \mathrm { AB } p$$
Since either $n - 1$ or $n$ has to be a multiple of $\mathbf { E }$ and either $n - 1 , n$ or $n + 1$ has to be a multiple of $\mathbf { F }$, and furthermore $\mathbf { E }$ and $\mathbf { F }$ are mutually prime, we know that $n ( n - 1 ) ( n + 1 )$ is a multiple of $\mathbf { G }$. (Write the answers in the order $1 < \square < \mathbf { E } < \mathbf { F } < \mathbf { G }$.) Hence one of $n - 1 , n$ and $n + 1$ must be a multiple of $\mathbf{HI}$.
So, since $n \leqq \mathrm { CD }$, the number of $n$ 's where $n - 1$ is a multiple of $\mathbf{HI}$ is $\mathbf{J}$, where $n$ is a multiple of $\mathbf{HI}$ is $\mathbf { K }$, and where $n + 1$ is a multiple of $\mathbf{HI}$ is $\mathbf { L }$.
Thus, the number of $n$ 's is $\mathbf { M N }$ and the prime numbers among them are $\mathbf { O P } , \mathbf { Q R }$, $\mathbf{ST}$, in ascending order.

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
$$x = \mathbf { R } , \quad y = \mathbf { S T } .$$

Let $N$ be a positive integer. Both when it is written in base 5 and when it is written in base 9, it is a 3-digit number, but the order of the numerals is reversed. We are to represent $N$ in base 10 (decimal) and in base 4.
Let $N$ be $abc$ in base 5 and $cba$ in base 9. Then we have
$$\mathbf { A } \leq a \leq \mathbf { B } , \quad \mathbf { C } \leq b \leq \mathbf { D } , \quad \mathbf { E } \leqq c \leqq \mathbf { F } \text {. }$$
Since we also have
$$N = \mathbf { G H } a + \mathbf { I } \quad b + c = \mathbf { J K } c + \mathbf { L } \quad b + a ,$$
we obtain
$$b = \mathbf { M } a - \mathbf { N O } c .$$
The $a$, $b$ and $c$ satisfying (1) and (2) are
$$a = \mathbf { P } , \quad b = \mathbf { Q } , \quad c = \mathbf { R } .$$
Thus $N$ expressed in base 10 is $\mathbf { S T U }$, and $N$ expressed in base 4 is $\mathbf { V W X Y }$.

The least common multiple of $b$ and $40$ is $120$.
Accordingly, how many different positive integers $b$ are there?
A) 6
B) 8
C) 10
D) 12
E) 14

For natural numbers $x$ and $y$
$$\begin{array} { r | r | r } x & \frac { 10 } { m } \\ { } ^ { - } & = ^ { y } \frac { 15 } { 3 } \end{array}$$
Given this, what is the remainder when the product $x \cdot y$ is divided by 5?
A) 0
B) 1
C) 2
D) 3
E) 4

On the set of positive integers, the operations $\oplus$ and $\otimes$ are defined using the greatest common divisor and least common multiple as follows:
$$\begin{aligned} & a \oplus b = \operatorname { GCD } ( a , b ) \\ & a \otimes b = \operatorname { LCM } ( a , b ) \end{aligned}$$
Accordingly, what is the result of the operation $18 \oplus ( 12 \otimes 4 )$?
A) 2
B) 3
C) 6
D) 8
E) 9

A two-digit number $AB$ is called a symmetric prime if both $AB$ and $BA$ are prime numbers.
For a symmetric prime number $AB$, which of the following cannot be the product A.B?
A) 7
B) 9
C) 15
D) 21
E) 63

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

$$\begin{aligned} & A = \left\{ n \in Z ^ { + } \mid n \leq 100 ; n \text{ is divisible by } 3 \right\} \\ & B = \left\{ n \in Z ^ { + } \mid n \leq 100 ; n \text{ is divisible by } 5 \right\} \end{aligned}$$
The sets are given. Accordingly, how many elements does the difference set $\mathrm { A } \backslash \mathrm { B }$ have?
A) 33
B) 32
C) 30
D) 28
E) 27

Let p and q be distinct prime numbers such that
$$\begin{aligned} & a = p ^ { 4 } \cdot q ^ { 2 } \\ & b = p ^ { 2 } \cdot q ^ { 3 } \end{aligned}$$
Which of the following is the greatest common divisor of numbers a and b?
A) $p ^ { 5 } \cdot q ^ { 4 }$
B) $p ^ { 4 } \cdot q ^ { 3 }$
C) $p ^ { 3 } \cdot q ^ { 4 }$
D) $p ^ { 2 } \cdot q ^ { 2 }$
E) $p ^ { 2 } \cdot q ^ { 3 }$

$$\begin{aligned} & 2 ^ { x } \equiv 1 ( \bmod 7 ) \\ & 3 ^ { y } \equiv 4 ( \bmod 7 ) \end{aligned}$$
For the smallest positive integers x and y satisfying these congruences, what is the difference $y - x$?
A) 5
B) 4
C) 3
D) 2
E) 1

Let a be a positive integer and $p = a^{2} + 5$. If p is a prime number, which of the following statements are true?
I. a is an even number. II. The remainder when p is divided by 4 is 1. III. $\mathrm{p} - 6$ is prime.
A) I and III B) Only I C) I and II D) Only III E) I, II and III

Let n be a positive integer, and let $S(n)$ denote the set of positive integers that divide n without remainder.
Accordingly, how many elements does the intersection set $S(60) \cap S(72)$ have?
A) 8 B) 9 C) 6 D) 5 E) 4

If a number of the form $7k + 4$ is divisible by 3 without remainder, how many positive integers k less than 21 are there?
A) 8 B) 9 C) 7 D) 6 E) 5

$$\prod _ { n = 1 } ^ { 7 } ( 3 n + 2 )$$
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

What is the representation in base 2 of the number $(15)_8$ given in base 8?
A) $(1001)_2$
B) $(1011)_2$
C) $(1101)_2$
D) $(1110)_2$
E) $(1111)_2$