exam: BalticWay

problem: A set of 8 problems was prepared for an examination. Each student was given 3 of them. No two students received more than one common problem. What is the largest possible number of students?

solution:
Answer: 8 .

Denote the problems by $A, B, C, D, E, F, G, H$, then 8 possible problem sets are $A B C, A D E, A F G, B D G, B F H, C D H, C E F, E G H$. Hence, there could be 8 students.

Suppose that some problem (e.g., $A$ ) was given to 4 students. Then each of these 4 students should receive 2 different "supplementary" problems, and there should be at least 9 problems - a contradiction. Therefore each problem was given to at most 3 students, and there were at most $8 \cdot 3=24$ "awardings" of problems. As each student was "awarded" 3 problems, there were at most 8 students.


problem: Let $n \geqslant 2$ be a positive integer. Find whether there exist $n$ pairwise nonintersecting nonempty subsets of $\{1,2,3, \ldots\}$ such that each positive integer can be expressed in a unique way as a sum of at most $n$ integers, all from different subsets.

solution:
Answer: yes.

Let $A_{1}$ be the set of positive integers whose only non-zero digits may be the 1 -st, the $(n+1)$-st, the $(2 n+1)$-st etc. from the end; $A_{2}$ be the set of positive integers whose only non-zero digits may be the 2 -nd, the $(n+2)$ nd, the $(2 n+2)$-nd etc. from the end, and so on. The sets $A_{1}, A_{2}, \ldots, A_{n}$ have the required property.

Remark. This problem is quite similar to problem 18 from Baltic Way 1997.


problem: The numbers $1,2, \ldots, 49$ are placed in a $7 \times 7$ array, and the sum of the numbers in each row and in each column is computed. Some of these 14 sums are odd while others are even. Let $A$ denote the sum of all the odd sums and $B$ the sum of all even sums. Is it possible that the numbers were placed in the array in such a way that $A=B$ ?

solution:
Answer: no.

If this were possible, then $2 \cdot(1+\ldots+49)=A+B=2 B$. But $B$ is even since it is the sum of even numbers, whereas $1+\ldots+49=25 \cdot 49$ is odd. This is a contradiction.


problem:
Let $p$ and $q$ be two different primes. Prove that

$$
\left\lfloor\frac{p}{q}\right\rfloor+\left\lfloor\frac{2 p}{q}\right\rfloor+\left\lfloor\frac{3 p}{q}\right\rfloor+\ldots+\left\lfloor\frac{(q-1) p}{q}\right\rfloor=\frac{1}{2}(p-1)(q-1) .
$$

(Here $\lfloor x\rfloor$ denotes the largest integer not greater than $x$.)

solution:
The line $y=\frac{p}{q} x$ contains the diagonal of the rectangle with vertices $(0,0)$, $(q, 0),(q, p)$ and $(0, p)$ and passes through no points with integer coordinates in the interior of that rectangle. For $k=1,2, \ldots, q-1$ the summand $\left\lfloor\frac{k p}{q}\right\rfloor$ counts the number of interior points of the rectangle lying below the diagonal $y=\frac{p}{q} x$ and having $x$-coordinate equal to $k$. Therefore the sum in consideration counts all interior points with integer coordinates below the diagonal, which is exactly half the number of all points with integer coordinates in the interior of the rectangle, i.e. $\frac{1}{2} \cdot(p-1)(q-1)$.

Remark. The integers $p$ and $q$ need not be primes: in the solution we only used the fact that they are coprime.


problem: Let 2001 given points on a circle be colored either red or green. In one step all points are recolored simultaneously in the following way: If both direct neighbors of a point $P$ have the same color as $P$, then the color of $P$ remains unchanged, otherwise $P$ obtains the other color. Starting with the first coloring $F_{1}$, we obtain the colorings $F_{2}, F_{3}, \ldots$ after several recoloring steps. Prove that there is a number $n_{0} \leqslant 1000$ such that $F_{n_{0}}=F_{n_{0}+2}$. Is the assertion also true if 1000 is replaced by 999 ?

solution:
Answer: no.

Let the points be denoted by $1,2, \ldots, 2001$ such that $i, j$ are neighbors if $|i-j|=1$ or $\{i, j\}=\{1,2001\}$. We say that $k$ points form a monochromatic segment of length $k$ if the points are consecutive on the circle and if
they all have the same color. For a coloring $F$ let $d(F)$ be the maximum length of a monochromatic segment. Note that $d\left(F_{n}\right)>1$ for all $n$ since 2001 is odd. If $d\left(F_{1}\right)=2001$ then all points have the same color, hence $F_{1}=F_{2}=F_{3}=\ldots$ and we can choose $n_{0}=1$. Thus, let $1<d\left(F_{1}\right)<2001$. Below we shall prove the following implications:

If $3<d\left(F_{n}\right)<2001$, then $d\left(F_{n+1}\right)=d\left(F_{n}\right)-2 ;$

If $d\left(F_{n}\right)=3$, then $d\left(F_{n+1}\right)=2$;

If $d\left(F_{n}\right)=2$, then $d\left(F_{n+1}\right)=d\left(F_{n}\right)$ and $F_{n+2}=F_{n}$;

From (1) and (2) it follows that $d\left(F_{1000}\right) \leqslant 2$, hence by (3) we have $F_{1000}=F_{1002}$. Moreover, if $F_{1}$ is the coloring where 1 is colored red and all other points are colored green, then $d\left(F_{1}\right)=2000$ and thus $d\left(F_{1}\right)>d\left(F_{2}\right)>\ldots>d\left(F_{1000}\right)=2$ which shows that, for all $n<1000, F_{n} \neq F_{n+2}$ and thus 1000 cannot be replaced by 999 .

It remains to prove (1)-(3). Let $(i+1, \ldots, i+k)$ be a longest monochromatic segment for $F_{n}$ (considering the labels of the points modulo 2001). Then $(i+2, \ldots, i+k-1)$ is a monochromatic segment for $F_{n+1}$ and thus $d\left(F_{n+1}\right) \geqslant d\left(F_{n}\right)-2$. Moreover, if $(i+1, \ldots, i+k)$ is a longest monochromatic segment for $F_{n+1}$ where $k \geqslant 3$, then $(i, \ldots, i+k+1)$ is a monochromatic segment for $F_{n}$. From this and $F_{n+1}>1$ the implications (1) and (2) clearly follow. For proof of (3) note that if $d\left(F_{n}\right) \leqslant 2$ then $F_{n+1}$ is obtained from $F_{n}$ by changing the colour of all points.

![](https://cdn.mathpix.com/cropped/2024_04_17_2e42d82bdb4a68411fc6g-05.jpg?height=462&width=522&top_left_y=1226&top_left_x=304)

Figure 1


problem: The points $A, B, C, D, E$ lie on the circle $c$ in this order and satisfy $A B \| E C$ and $A C \| E D$. The line tangent to the circle $c$ at $E$ meets the line $A B$ at $P$. The lines $B D$ and $E C$ meet at $Q$. Prove that $|A C|=|P Q|$.

solution:
The arcs $B C$ and $A E$ are of equal length (see Figure 1). Also, since
$A B \| E C$ and $E D \| A C$, we have $\angle C A B=\angle D E C$ and the $\operatorname{arcs} D C$ and $B C$ are of equal length. Since $P E$ is tangent to $c$ and $|A E|=|D C|$, then $\angle P E A=\angle D B C=\angle Q B C$. As $A B C D$ is inscribed in $c$, we have $\angle Q C B=180^{\circ}-\angle E A B=\angle P A E$. Also, $A B C D$ is an isosceles trapezium, whence $|A E|=|B C|$. So the triangles $A P E$ and $C Q B$ are congruent, and $|Q C|=|P A|$. Now $P A C Q$ is a quadrilateral with a pair of opposite sides equal and parallel. So $P A C Q$ is a parallelogram, and $|P Q|=|A C|$.


problem:
Given a parallelogram $A B C D$. A circle passing through $A$ meets the line segments $A B, A C$ and $A D$ at inner points $M, K, N$, respectively. Prove that

$$
|A B| \cdot|A M|+|A D| \cdot|A N|=|A K| \cdot|A C| .
$$

solution:
Let $X$ be the point on segment $A C$ such that $\angle A D X=\angle A K N$, then

$$
\angle A X D=\angle A N K=180^{\circ}-\angle A M K
$$

(see Figure 2). Triangles $N A K$ and $X A D$ are similar, having two pairs of equal angles, hence $|A X|=\frac{|A N| \cdot|A D|}{|A K|}$. Since triangles $M A K$ and $X C D$ are also similar, we have $|C X|=\frac{|A M| \cdot|C D|}{|A K|}=\frac{|A M| \cdot|A B|}{|A K|}$ and

$$
|A M| \cdot|A B|+|A N| \cdot|A D|=(|A X|+|C X|) \cdot|A K|=|A C| \cdot|A K| \text {. }
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_2e42d82bdb4a68411fc6g-06.jpg?height=408&width=783&top_left_y=1031&top_left_x=319)

Figure 2


problem:
Let $A B C D$ be a convex quadrilateral, and let $N$ be the midpoint of $B C$. Suppose further that $\angle A N D=135^{\circ}$. Prove that

$$
|A B|+|C D|+\frac{1}{\sqrt{2}} \cdot|B C| \geqslant|A D|
$$

solution:
Let $X$ be the point symmetric to $B$ with respect to $A N$, and let $Y$ be the point symmetric to $C$ with respect to $D N$ (see Figure 3 ). Then

$$
\angle X N Y=180^{\circ}-2 \cdot\left(180^{\circ}-135^{\circ}\right)=90^{\circ}
$$

and $|N X|=|N Y|=\frac{|B C|}{2}$. Therefore, $|X Y|=\frac{|B C|}{\sqrt{2}}$. Moreover, we have

$$
\begin{aligned}
& |A X|=|A B| \text { and }|D Y|=|D C| \text {. Consequently, } \\
& \qquad|A D| \leqslant|A X|+|X Y|+|Y D|=|A B|+\frac{|B C|}{\sqrt{2}}+|D C| .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_2e42d82bdb4a68411fc6g-07.jpg?height=214&width=700&top_left_y=397&top_left_x=255)

Figure 3


problem: Given a rhombus $A B C D$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle A P D+\angle B P C=180^{\circ}$.

solution:
Answer: the locus of the points $P$ is the union of the diagonals $A C$ and $B D$.

Let $Q$ be a point such that $P Q C D$ is a parallelogram (see Figure 4). Then $A B Q P$ is also a parallelogram. From the equality $\angle A P D+\angle B P C=180^{\circ}$ it follows that $\angle B Q C+\angle B P C=180^{\circ}$, so the points $B, Q, C, P$ lie on a common circle. Therefore, $\angle P B C=\angle P Q C=\angle P D C$, and since $|B C|=|C D|$, we obtain that $\angle C P B=\angle C P D$ or $\angle C P B+\angle C P D=180^{\circ}$. Hence, the point $P$ lies on the segment $A C$ or on the segment $B D$.
![](https://cdn.mathpix.com/cropped/2024_04_17_2e42d82bdb4a68411fc6g-07.jpg?height=294&width=1076&top_left_y=1062&top_left_x=203)

Figure 4

Conversely, any point $P$ lying on the diagonal $A C$ satisfies the equation $\angle B P C=\angle D P C$. Therefore, $\angle A P D+\angle B P C=180^{\circ}$. Analogously, we show that the last equation holds if the point $P$ lies on the diagonal $B D$.


problem: In a triangle $A B C$, the bisector of $\angle B A C$ meets the side $B C$ at the point $D$. Knowing that $|B D| \cdot|C D|=|A D|^{2}$ and $\angle A D B=45^{\circ}$, determine the angles of triangle $A B C$.

solution:
Answer: $\angle B A C=60^{\circ}, \angle A B C=105^{\circ}$ and $\angle A C B=15^{\circ}$.

Suppose the line $A D$ meets the circumcircle of triangle $A B C$ at $A$ and $E$ (see Figure 5). Let $M$ be the midpoint of $B C$ and $O$ the circumcentre of triangle $A B C$. Since the arcs $B E$ and $E C$ are equal, then the points $O$, $M, E$ are collinear and $O E$ is perpendicular to $B C$. From the equality
$\angle C D E=\angle A D B=45^{\circ}$ it follows that $\angle A E O=45^{\circ}$. Since $|A O|=|E O|$, we have $\angle A O E=90^{\circ}$ and $A O \| D M$.

From the equality $|B D| \cdot|C D|=|A D|^{2}$ we obtain $|A D|=|D E|$, which implies that $|O M|=|M E|$. Therefore $|B O|=|B E|$ and also $|B O|=|E O|$. Hence the triangle $B O E$ is equilateral. This gives $\angle B A E=30^{\circ}$, so $\angle B A C=60^{\circ}$. Summing up the angles of the triangle $A B D$ we obtain $\angle A B C=105^{\circ}$ and from this $\angle A C B=15^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2e42d82bdb4a68411fc6g-08.jpg?height=374&width=458&top_left_y=515&top_left_x=316)

Figure 5


problem:
The real-valued function $f$ is defined for all positive integers. For any integers $a>1, b>1$ with $d=\operatorname{gcd}(a, b)$, we have

$$
f(a b)=f(d) \cdot\left(f\left(\frac{a}{d}\right)+f\left(\frac{b}{d}\right)\right),
$$

Determine all possible values of $f(2001)$.

solution:
Answer: 0 and $\frac{1}{2}$.

Obviously the constant functions $f(n)=0$ and $f(n)=\frac{1}{2}$ provide solutions.

We show that there are no other solutions. Assume $f(2001) \neq 0$. Since $2001=3 \cdot 667$ and $\operatorname{gcd}(3,667)=1$, then

$$
f(2001)=f(1) \cdot(f(3)+f(667)),
$$

and $f(1) \neq 0$. Since $\operatorname{gcd}(2001,2001)=2001$ then

$$
f\left(2001^{2}\right)=f(2001)(2 \cdot f(1)) \neq 0 .
$$

Also $\operatorname{gcd}\left(2001,2001^{3}\right)=2001$, so

$$
f\left(2001^{4}\right)=f(2001) \cdot\left(f(1)+f\left(2001^{2}\right)\right)=f(1) f(2001)(1+2 f(2001))
$$

On the other hand, $\operatorname{gcd}\left(2001^{2}, 2001^{2}\right)=2001^{2}$ and

$$
f\left(2001^{4}\right)=f\left(2001^{2}\right) \cdot(f(1)+f(1))=2 f(1) f\left(2001^{2}\right)=4 f(1)^{2} f(2001) .
$$

So $4 f(1)=1+2 f(2001)$ and $f(2001)=2 f(1)-\frac{1}{2}$. Exactly the same
argument starting from $f\left(2001^{2}\right) \neq 0$ instead of $f(2001)$ shows that $f\left(2001^{2}\right)=2 f(1)-\frac{1}{2}$. So

$$
2 f(1)-\frac{1}{2}=2 f(1)\left(2 f(1)-\frac{1}{2}\right) \text {. }
$$

Since $2 f(1)-\frac{1}{2}=f(2001) \neq 0$, we have $f(1)=\frac{1}{2}$, which implies $f(2001)=2 f(1)-\frac{1}{2}=\frac{1}{2}$.


solution:
By Hölder's inequality,

$$
\sum_{i=1}^{n} a^{3}=\sum_{i=1}^{n}\left(a_{i} \cdot a_{i}^{2}\right) \leqslant\left(\sum_{i=1}^{n} a_{i}^{5 / 3}\right)^{3 / 5} \cdot\left(\sum_{i=1}^{n}\left(a_{i}^{2}\right)^{5 / 2}\right)^{2 / 5} .
$$

We will show that

$$
\sum_{i=1}^{n} a_{i}^{5 / 3} \leqslant\left(\sum_{i=1}^{n} a_{i}\right)^{5 / 3}
$$

Let $S=\sum_{i=1}^{n} a_{i}$, then (4) is equivalent to

$$
\sum_{i=1}^{n}\left(\frac{a_{i}}{S}\right)^{5 / 3} \leqslant 1=\sum_{i=1}^{n} \frac{a_{i}}{S}
$$

which holds since $0<\frac{a_{i}}{S} \leqslant 1$ and $\frac{5}{3}>1$ yield $\left(\frac{a_{i}}{S}\right)^{5 / 3} \leqslant \frac{a_{i}}{S}$. So,

$$
\sum_{i=1}^{n} a_{i}^{3} \leqslant\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} a_{i}^{5}\right)^{2 / 5}
$$

which gives $\sum_{i=1}^{n} a_{i} \geqslant \frac{3}{5^{2 / 5}}>\frac{3}{2}$, since $2^{5}>5^{2}$ and hence $2>5^{2 / 5}$.


problem:
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of real numbers satisfying $a_{0}=1$ and $a_{n}=a_{\lfloor 7 n / 9\rfloor}+a_{\lfloor n / 9\rfloor}$ for $n=1,2, \ldots$ Prove that there exists a positive integer $k$ with $a_{k}<\frac{k}{2001 !}$.

(Here $\lfloor x\rfloor$ denotes the largest integer not greater than $x$.)

solution:
Consider the equation

$$
\left(\frac{7}{9}\right)^{x}+\left(\frac{1}{9}\right)^{x}=1 .
$$

It has a root $\frac{1}{2}<\alpha<1$, because $\sqrt{\frac{7}{9}}+\sqrt{\frac{1}{9}}=\frac{\sqrt{7}+1}{3}>1$ and $\frac{7}{9}+\frac{1}{9}<1$. We will prove that $a_{n} \leqslant M \cdot n^{\alpha}$ for some $M>0-$ since $\frac{n^{\alpha}}{n}$ will be arbitrarily small for large enough $n$, the claim follows from this immediately. We choose $M$ so that the inequality $a_{n} \leqslant M \cdot n^{\alpha}$ holds for $1 \leqslant n \leqslant 8$; since for $n \geqslant 9$ we have $1<[7 n / 9]<n$ and $1 \leqslant[n / 9]<n$, it follows by induction that

$$
\begin{aligned}
a_{n} & =a_{[7 n / 9]}+a_{[n / 9]} \leqslant M \cdot\left[\frac{7 n}{9}\right]^{\alpha}+M \cdot\left[\frac{n}{9}\right]^{\alpha} \leqslant \\
& \leqslant M \cdot\left(\frac{7 n}{9}\right)^{\alpha}+M \cdot\left(\frac{n}{9}\right)^{\alpha}=M \cdot n^{\alpha} \cdot\left(\left(\frac{7}{9}\right)^{\alpha}+\left(\frac{1}{9}\right)^{\alpha}\right)=M \cdot n^{\alpha} .
\end{aligned}
$$


problem:
There are $2 n$ cards. On each card some real number $x, 1 \leqslant x \leqslant 2$, is written (there can be different numbers on different cards). Prove that
the cards can be divided into two heaps with sums $s_{1}$ and $s_{2}$ so that $\frac{n}{n+1} \leqslant \frac{s_{1}}{s_{2}} \leqslant 1$.

solution:
Let the numbers be $x_{1} \leqslant x_{2} \leqslant \ldots \leqslant x_{2 n-1} \leqslant x_{2 n}$. We will show that the choice $s_{1}=x_{1}+x_{3}+x_{5}+\cdots+x_{2 n-1}$ and $s_{2}=x_{2}+x_{4}+\cdots+x_{2 n}$ solves the problem. Indeed, the inequality $\frac{s_{1}}{s_{2}} \leqslant 1$ is obvious and we have

$$
\begin{aligned}
\frac{s_{1}}{s_{2}} & =\frac{x_{1}+x_{3}+x_{5}+\ldots+x_{2 n-1}}{x_{2}+x_{4}+x_{6}+\ldots+x_{2 n}}=\frac{\left(x_{3}+x_{5}+\ldots+x_{2 n-1}\right)+x_{1}}{\left(x_{2}+x_{4}+\ldots+x_{2 n-2}\right)+x_{2 n}} \geqslant \\
& \geqslant \frac{\left(x_{3}+x_{5}+\ldots+x_{2 n-1}\right)+1}{\left(x_{2}+x_{4}+\ldots+x_{2 n-2}\right)+2} \geqslant \frac{\left(x_{2}+x_{4}+\ldots+x_{2 n-2}\right)+1}{\left(x_{2}+x_{4}+\ldots+x_{2 n-2}\right)+2}= \\
& =1-\frac{1}{\left(x_{2}+x_{4}+\ldots+x_{2 n-2}\right)+2} \geqslant 1-\frac{1}{(n-1)+2}=\frac{n}{n+1} .
\end{aligned}
$$


problem: Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive real numbers satisfying $i \cdot a_{i}^{2} \geqslant(i+1) \cdot a_{i-1} a_{i+1}$ for $i=1,2, \ldots$ Furthermore, let $x$ and $y$ be positive reals, and let $b_{i}=x a_{i}+y a_{i-1}$ for $i=1,2, \ldots$ Prove that the inequality $i \cdot b_{i}^{2}>(i+1) \cdot b_{i-1} b_{i+1}$ holds for all integers $i \geqslant 2$.

solution:
Let $i \geqslant 2$. We are given the inequalities

$$
(i-1) \cdot a_{i-1}^{2} \geqslant i \cdot a_{i} a_{i-2}
$$

and

$$
i \cdot a_{i}^{2} \geqslant(i+1) \cdot a_{i+1} a_{i-1} .
$$

Multiplying both sides of (6) by $x^{2}$, we obtain

$$
i \cdot x^{2} \cdot a_{i}^{2} \geqslant(i+1) \cdot x^{2} \cdot a_{i+1} a_{i-1}
$$

By (5),

$$
\frac{a_{i-1}^{2}}{a_{i} a_{i-2}} \geqslant \frac{i}{i-1}=1+\frac{1}{i-1}>1+\frac{1}{i}=\frac{i+1}{i}
$$

which implies

$$
i \cdot y^{2} \cdot a_{i-1}^{2}>(i+1) \cdot y^{2} \cdot a_{i} a_{i-2} .
$$

Multiplying (5) and (6), and dividing both sides of the resulting inequality by $i a_{i} a_{i-1}$, we get

$$
(i-1) \cdot a_{i} a_{i-1} \geqslant(i+1) \cdot a_{i+1} a_{i-2} .
$$

Adding $(i+1) a_{i} a_{i-1}$ to both sides of the last inequality and multiplying both sides of the resulting inequality by $x y$ gives

$$
i \cdot 2 x y \cdot a_{i} a_{i-1} \geqslant(i+1) \cdot x y \cdot\left(a_{i+1} a_{i-2}+a_{i} a_{i-1}\right) .
$$

Finally, adding up (7), (8) and (9) results in

$$
i \cdot\left(x a_{i}+y a_{i-1}\right)^{2}>(i+1) \cdot\left(x a_{i+1}+y a_{i}\right)\left(x a_{i-1}+y a_{i-2}\right)
$$

which is equivalent to the claim.


problem:
Let $f$ be a real-valued function defined on the positive integers satisfying the following condition: For all $n>1$ there exists a prime divisor $p$ of $n$ such that

$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$

Given that $f(2001)=1$, what is the value of $f(2002)$ ?

solution:
Answer: 2.

For any prime $p$ we have $f(p)=f(1)-f(p)$ and thus $f(p)=\frac{f(1)}{2}$. If $n$ is a product of two primes $p$ and $q$, then $f(n)=f(p)-f(q)$ or $f(n)=f(q)-f(p)$, so $f(n)=0$. By the same reasoning we find that if $n$ is a product of three primes, then there is a prime $p$ such that

$$
f(n)=f\left(\frac{n}{p}\right)-f(p)=-f(p)=-\frac{f(1)}{2} .
$$

By simple induction we can show that if $n$ is the product of $k$ primes, then $f(n)=(2-k) \cdot \frac{f(1)}{2}$. In particular, $f(2001)=f(3 \cdot 23 \cdot 29)=1$ so $f(1)=-2$. Therefore, $f(2002)=f(2 \cdot 7 \cdot 11 \cdot 13)=-f(1)=2$.


problem: Let $n$ be a positive integer. Prove that at least $2^{n-1}+n$ numbers can be chosen from the set $\left\{1,2,3, \ldots, 2^{n}\right\}$ such that for any two different chosen numbers $x$ and $y, x+y$ is not a divisor of $x \cdot y$.

solution:
We choose the numbers $1,3,5, \ldots, 2^{n}-1$ and $2,4,8,16, \ldots, 2^{n}$, i.e. all odd numbers and all powers of 2 . Consider the three possible cases.

(1) If $x=2 a-1$ and $y=2 b-1$, then $x+y=(2 a-1)+(2 b-1)=2(a+b-1)$ is even and does not divide $x y=(2 a-1)(2 b-1)$ which is odd.

(2) If $x=2^{k}$ and $y=2^{m}$ where $k<m$, then $x+y=2^{k}\left(2^{m-k}+1\right)$ has an odd divisor greater than 1 and hence does not divide $x y=2^{a+b}$.

(3) If $x=2^{k}$ and $y=2 b-1$, then $x+y=2^{k}+(2 b-1)>(2 b-1)$ is odd and hence does not divide $x y=2^{k}(2 b-1)$ which has $2 b-1$ as its largest odd divisor.


problem: Let $a$ be an odd integer. Prove that $a^{2^{n}}+2^{2^{n}}$ and $a^{2^{m}}+2^{2^{m}}$ are relatively prime for all positive integers $n$ and $m$ with $n \neq m$.

solution:
Rewriting $a^{2^{n}}+2^{2^{n}}=a^{2^{n}}-2^{2^{n}}+2 \cdot 2^{2^{n}}$ and making repeated use of the identity

$$
a^{2^{n}}-2^{2^{n}}=\left(a^{2^{n-1}}-2^{2^{n-1}}\right) \cdot\left(a^{2^{n-1}}+2^{2^{n-1}}\right)
$$

we get

$$
\begin{gathered}
a^{2^{n}}+2^{2^{n}}=\left(a^{2^{n-1}}+2^{2^{n-1}}\right) \cdot\left(a^{2^{n-2}}+2^{2^{n-2}}\right) \cdot \ldots \cdot\left(a^{2^{m}}+2^{2^{m}}\right) \cdot \ldots \\
\ldots \cdot\left(a^{2}+2^{2}\right) \cdot(a+2) \cdot(a-2)+2 \cdot 2^{2^{n}}
\end{gathered}
$$

For $n>m$, assume that $a^{2^{n}}+2^{2^{n}}$ and $a^{2^{m}}+2^{2^{m}}$ have a common divisor $d>1$. Then an odd integer $d$ divides $2 \cdot 2^{2^{n}}$, a contradiction.


problem: What is the smallest positive odd integer having the same number of positive divisors as 360 ?

solution:
Answer: 31185 .

An integer with the prime factorization $p_{1}^{r_{1}} \cdot p_{2}^{r_{2}} \cdot \ldots \cdot p_{k}^{r_{k}}$ (where $p_{1}, p_{2}, \ldots$, $p_{k}$ are distinct primes) has precisely $\left(r_{1}+1\right) \cdot\left(r_{2}+1\right) \cdot \ldots \cdot\left(r_{k}+1\right)$ distinct positive divisors. Since $360=2^{3} \cdot 3^{2} \cdot 5$, it follows that 360 has $4 \cdot 3 \cdot 2=24$ positive divisors. Since $24=3 \cdot 2 \cdot 2 \cdot 2$, it is easy to check that the smallest odd number with 24 positive divisors is $3^{2} \cdot 5 \cdot 7 \cdot 11=31185$.


problem:
From a sequence of integers $(a, b, c, d)$ each of the sequences

$$
(c, d, a, b),(b, a, d, c),(a+n c, b+n d, c, d),(a+n b, b, c+n d, d),
$$

for arbitrary integer $n$ can be obtained by one step. Is it possible to obtain $(3,4,5,7)$ from $(1,2,3,4)$ through a sequence of such steps?

## Solutions

solution:
Answer: no.

Under all transformations $(a, b, c, d) \rightarrow\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)$ allowed in the problem we have $|a d-b c|=\left|a^{\prime} d^{\prime}-b^{\prime} c^{\prime}\right|$, but $|1 \cdot 4-2 \cdot 3|=2 \neq 1=|3 \cdot 7-4 \cdot 5|$.

Remark. The transformations allowed in the problem are in fact the elementary transformations of the determinant

$$
\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|
$$

and the invariant $|a d-b c|$ is the absolute value of the determinant which is preserved under these transformations.