BalticWay/md/en-bw16sol.md

Baltic Way 2016 - Solutions

1. Find all pairs of primes $(p, q)$ such that

$$p^3-q^5=(p+q{)}^2\mathrm{.}p^\{3\}-q^\{5\}=(p+q)^\{2\}\; .$$

Solution. Assume first that neither of the numbers equals 3. Then, if $p \equiv q \bmod 3$, the left hand side is divisible by 3 , but the right hand side is not. But if $p \equiv-q \bmod 3$, the left hand side is not divisible by 3 , while the right hand side is. So this is not possible.

If $p=3$, then $q^{5}<27$, which is impossible. Therefore $q=3$, and the equation turns into $p^{3}-243=(p+3)^{2}$ or

$$p(p^2-p-6)=252=7\cdot 36\text{. }p\backslash left(p^\{2\}-p-6\backslash right)=252=7\; \backslash cdot\; 36\; \backslash text\; \{.\; \}$$

As $p>3$ then $p^{2}-p-6$ is positive and increases with $p$. So the equation has at most one solution. It is easy to see that $p=7$ is the one and $(7,3)$ is a solution to the given equation.

2. Prove or disprove the following hypotheses.

a) For all $k \geq 2$, each sequence of $k$ consecutive positive integers contains a number that is not divisible by any prime number less than $k$.

b) For all $k \geq 2$, each sequence of $k$ consecutive positive integers contains a number that is relatively prime to all other members of the sequence.

Solution. We give a counterexample to both claims. So neither of them is true.

For a), a counterexample is the sequence $(2,3,4,5,6,7,8,9)$ of eight consecutive integers all of which are divisible by some prime less than 8 .

To construct a counterexample to b), we notice that by the Chinese Remainder Theorem, there exists an integer $x$ such that $x \equiv 0 \bmod 2, x \equiv 0 \bmod 5, x \equiv 0 \bmod 11, x \equiv 2 \bmod 3$, $x \equiv 5 \bmod 7$ and $x \equiv 10 \bmod 13$. The last three of these congruences mean that $x+16$ is a multiple of 3,7 , and 13 . Now consider the sequence $(x, x+1, \ldots, x+16)$ of 17 consequtive integers. Of these all numbers $x+2 k, 0 \leq k \leq 8$, are even and so have a common factor with some other. Of the remaining, $x+1, x+7$ and $x+13$ are divisible by $3, x+3$ is a multiple of 13 as is $x+16, x+5$ is divisible by 5 as $x, x+9$ is a multiple of 7 as $x+2$, $x+11$ a multiple of 11 as is $x$, and finally $x+15$ is a multiple of 5 as is $x$.

Remark. The counterexample given to either hypothesis is the shortest possible. The only counterexamples of length 8 to the first hypothesis are those where numbers give remainders $2,3, \ldots, 9 ; 3,4, \ldots, 10 ;-2,-3, \ldots,-9 ;$ or $-3,-4, \ldots,-10$ modulo 210 . The only counterexamples of length 17 to the second hypothesis are those where the numbers give remainders $2184,2185, \ldots, 2200$ or $-2184,-2185, \ldots,-2200$ modulo 30030.

3. For which integers $n=1, \ldots, 6$ does the equation

$$a^n+b^n=c^n+na^\{n\}+b^\{n\}=c^\{n\}+n$$

have a solution in integers?

Solution. A solution clearly exists for $n=1,2,3$ :

$$1^1+0^1=0^1+1,1^2+1^2=0^2+2,1^3+1^3=(-1{)}^3+3.1^\{1\}+0^\{1\}=0^\{1\}+1,\; \backslash quad\; 1^\{2\}+1^\{2\}=0^\{2\}+2,\; \backslash quad\; 1^\{3\}+1^\{3\}=(-1)^\{3\}+3\; .$$

We show that for $n=4,5,6$ there is no solution.

For $n=4$, the equation $a^{4}+b^{4}=c^{4}+4$ may be considered modulo 8 . Since each fourth power $x^{4} \equiv 0,1 \bmod 8$, the expression $a^{4}+b^{4}-c^{4}$ can never be congruent to 4 .

For $n=5$, consider the equation $a^{5}+b^{5}=c^{5}+5$ modulo 11 . As $x^{5} \equiv 0$ or $\equiv \pm 1 \bmod 11$ (This can be seen by Fermat's Little Theorem or by direct computation), $a^{5}+b^{5}-c^{5}$ cannot be congruent to 5 .

The case $n=6$ is similarly dismissed by considering the equation modulo 13 .

4. Let $n$ be a positive integer and let $a, b, c, d$ be integers such that $n \mid a+b+c+d$ and $n \mid a^{2}+b^{2}+c^{2}+d^{2}$. Show that

$$n\mid a^4+b^4+c^4+d^4+4abcd\mathrm{.}n\; \backslash mid\; a^\{4\}+b^\{4\}+c^\{4\}+d^\{4\}+4\; a\; b\; c\; d\; .$$

Solution 1. Consider the polynomial

$$w(x)=(x-a)(x-b)(x-c)(x-d)=x^4+A{x}^3+B{x}^2+Cx+D\mathrm{.}w(x)=(x-a)(x-b)(x-c)(x-d)=x^\{4\}+A\; x^\{3\}+B\; x^\{2\}+C\; x+D\; .$$

It is clear that $w(a)=w(b)=w(c)=w(d)=0$. By adding these values we get

$$\begin{array}{c}{\displaystyle w(a)+w(b)+w(c)+w(d)=a^4+b^4+c^4+d^4+A(a^3+b^3+c^3+d^3)+}\\ {\displaystyle +B(a^2+b^2+c^2+d^2)+C(a+b+c+d)+4D=0.}\end{array}\backslash begin\{gathered\}\; w(a)+w(b)+w(c)+w(d)=a^\{4\}+b^\{4\}+c^\{4\}+d^\{4\}+A\backslash left(a^\{3\}+b^\{3\}+c^\{3\}+d^\{3\}\backslash right)+\; \backslash \backslash \; +B\backslash left(a^\{2\}+b^\{2\}+c^\{2\}+d^\{2\}\backslash right)+C(a+b+c+d)+4\; D=0\; .\; \backslash end\{gathered\}$$

Hence

$$\begin{array}{c}{\displaystyle a^4+b^4+c^4+d^4+4D}\\ {\displaystyle =-A(a^3+b^3+c^3+d^3)-B(a^2+b^2+c^2+d^2)-C(a+b+c+d)\mathrm{.}}\end{array}\backslash begin\{gathered\}\; a^\{4\}+b^\{4\}+c^\{4\}+d^\{4\}+4\; D\; \backslash \backslash \; =-A\backslash left(a^\{3\}+b^\{3\}+c^\{3\}+d^\{3\}\backslash right)-B\backslash left(a^\{2\}+b^\{2\}+c^\{2\}+d^\{2\}\backslash right)-C(a+b+c+d)\; .\; \backslash end\{gathered\}$$

Using Vieta's formulas, we can see that $D=a b c d$ and $-A=a+b+c+d$. Therefore the right hand side of the equation above is divisible by $n$, and so is the left hand side.

Solution 2. Since the numbers $(a+b+c+d)\left(a^{3}+b^{3}+c^{3}+d^{3}\right),\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(a b+$ $a c+a d+b c+b d+c d)$ and $(a+b+c+d)(a b c+a c d+a b d+b c d)$ are divisible by $n$, then so is the number

$$\begin{array}{c}{\displaystyle (a+b+c+d)(a^3+b^3+c^3+d^3)-(a^2+b^2+c^2+d^2)(ab+ac+ad+bc+bd+cd)+}\\ {\displaystyle +(a+b+c+d)(abc+acd+abd+bcd)=a^4+b^4+c^4+d^4+4abcd\mathrm{.}}\end{array}\backslash begin\{gathered\}\; (a+b+c+d)\backslash left(a^\{3\}+b^\{3\}+c^\{3\}+d^\{3\}\backslash right)-\backslash left(a^\{2\}+b^\{2\}+c^\{2\}+d^\{2\}\backslash right)(a\; b+a\; c+a\; d+b\; c+b\; d+c\; d)+\; \backslash \backslash \; +(a+b+c+d)(a\; b\; c+a\; c\; d+a\; b\; d+b\; c\; d)=a^\{4\}+b^\{4\}+c^\{4\}+d^\{4\}+4\; a\; b\; c\; d\; .\; \backslash end\{gathered\}$$

(Heiki Niglas, Estonia)

5. Let $p>3$ be a prime such that $p \equiv 3(\bmod 4)$. Given a positive integer $a_{0}$, define the sequence $a_{0}, a_{1}, \ldots$ of integers by $a_{n}=a_{n-1}^{2^{n}}$ for all $n=1,2, \ldots$ Prove that it is possible to choose $a_{0}$ such that the subsequence $a_{N}, a_{N+1}, a_{N+2}, \ldots$ is not constant modulo $p$ for any positive integer $N$.

Solution. Let $p$ be a prime with residue 3 modulo 4 and $p>3$. Then $p-1=u \cdot 2$ where $u>1$ is odd. Choose $a_{0}=2$. The order of 2 modulo $p$ (that is, the smallest positive integer $t$ such that $\left.2^{t} \equiv 1 \bmod p\right)$ is a divisor of $\phi(p)=p-1=u \cdot 2$, but not a divisor of 2 since $1<2^{2}<p$. Hence the order of 2 modulo $p$ is not a power of 2 . By definition we see that $a_{n}=a_{0}^{2^{1+2+\cdots+n}}$. Since the order of $a_{0}=2$ modulo $p$ is not a power of 2 , we know that $a_{n} \not \equiv 1 \quad(\bmod p)$ for all $n=1,2,3, \ldots$ We proof the statement by contradiction. Assume there exists a positive integer $N$ such that $a_{n} \equiv a_{N}(\bmod p)$ for all $n \geq N$. Let $d>1$ be the order of $a_{N}$ modulo $p$. Then $a_{N} \equiv a_{n} \equiv a_{n+1}=a_{n}^{2^{n+1}} \equiv a_{N}^{2^{n+1}} \quad(\bmod p)$, and hence $a_{N}^{2^{n+1}-1} \equiv 1 \quad(\bmod p)$ for all $n \geq N$. Now $d$ divides $2^{n+1}-1$ for all $n \geq N$, but this is a contradiction since

$\operatorname{gcd}\left(2^{n+1}-1,2^{n+2}-1\right)=\operatorname{gcd}\left(2^{n+1}-1,2^{n+2}-1-2\left(2^{n+1}-1\right)\right)=\operatorname{gcd}\left(2^{n+1}-1,1\right)=1$.

Hence there does not exist such an $N$.

6. The set ${1,2, \ldots, 10}$ is partitioned into three subsets $A, B$ and $C$. For each subset the sum of its elements, the product of its elements and the sum of the digits of all its elements are calculated. Is it possible that $A$ alone has the largest sum of elements, $B$ alone has the largest product of elements, and $C$ alone has the largest sum of digits?

Solution. It is indeed possible. Choose $A={1,9,10}, B={3,7,8}, C={2,4,5,6}$. Then the sum of elements in $A, B$ and $C$, respectively, is 20,18 and 17 , the sum of digits 11, 18 and 17, while the product of elements is 90,168 and 240.

7. Find all positive integers $n$ for which

$$3x^n+n(x+2)-3\ge n{x}^{2}3\; x^\{n\}+n(x+2)-3\; \backslash geq\; n\; x^\{2\}$$

holds for all real numbers $x$.

Solution. We show that the inequality holds for even $n$ and only for them.

If $n$ is odd, the for $x=-1$ the left hand side of the inequality equals $n-6$ while the right hand side is $n$. So the inequality is not true for $x=-1$ for any odd $n$. So now assume that $n$ is even. Since $|x| \geq x$, it is enough to prove $3 x^{n}+2 n-3 \geq n x^{2}+n|x|$ for all $x$ or equivalently that $3 x^{n}+(2 n-3) \geq n x^{2}+n x$ for $x \geq 0$. Now the AGM-inequality gives

$$2x^n+(n-2)=x^n+x^n+1+\cdots +1\ge n{(x^n\cdot x^n\cdot 1^{n-2})}^{\frac{1}{n}}=n{x}^2\text{, }2\; x^\{n\}+(n-2)=x^\{n\}+x^\{n\}+1+\backslash cdots+1\; \backslash geq\; n\backslash left(x^\{n\}\; \backslash cdot\; x^\{n\}\; \backslash cdot\; 1^\{n-2\}\backslash right)^\{\backslash frac\{1\}\{n\}\}=n\; x^\{2\}\; \backslash text\; \{,\; \}$$

and similarly

$$x^n+(n-1)\ge n{(x^n\cdot 1^{n-1})}^{\frac{1}{n}}=nx\text{. }x^\{n\}+(n-1)\; \backslash geq\; n\backslash left(x^\{n\}\; \backslash cdot\; 1^\{n-1\}\backslash right)^\{\backslash frac\{1\}\{n\}\}=n\; x\; \backslash text\; \{.\; \}$$

Adding (1) and (2) yields the claim.

8. Find all real numbers a for which there exists a non-constant function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the following two equations for all $x \in \mathbb{R}$ : i) $\quad f(a x)=a^{2} f(x)$ and

ii) $f(f(x))=a f(x)$.

Solution. The conditions of the problem give two representations for $f(f(f(x)))$ :

$$f(f(f(x)))=af(f(x))=a^{2}f(x)f(f(f(x)))=a\; f(f(x))=a^\{2\}\; f(x)$$

and

$$f(f(f(x)))=f(af(x))=a^{2}f(f(x))=a^{3}f(x)\mathrm{.}f(f(f(x)))=f(a\; f(x))=a^\{2\}\; f(f(x))=a^\{3\}\; f(x)\; .$$

So $a^{2} f(x)=a^{3} f(x)$ for all $x$, and if there is an $x$ such that $f(x) \neq 0$, the $a=0$ or $a=1$. Otherwise $f$ is the constant function $f(x)=0$ for all $x$. If $a=1$, the function $f(x)=x$ satisfies the conditions. For $a=0$, one possible solution is the function $f$,

9. Find all quadruples $(a, b, c, d)$ of real numbers that simultaneously satisfy the following equations:

Solution 1. Consider the polynomial $P(x)=(a x+b)^{3}+(c x+d)^{3}=\left(a^{3}+b^{3}\right) x^{3}+3\left(a^{2} b+\right.$ $\left.c^{2} d\right) x^{2}+3\left(a b^{2}+c d^{2}\right) x+b^{3}+d^{3}$. By the conditions of the problem, $P(x)=2 x^{3}-18 x+1$. Clearly $P(0)>0, P(1)<0$ and $P(3)>0$. Thus $P$ has three distinct zeroes. But $P(x)=0$ implies $a x+b=-(c x+d)$ or $(a+c) x+b+d=0$. This equation has only one solution, unless $a=-c$ and $b=-d$. But since the conditions of the problem do not allow this, we infer that the system of equations in the problem has no solution.

Solution 2. If $0 \in{a, b}$, then one easily gets that $0 \in{c, d}$, which contradicts the equation $a b^{2}+c d^{2}=-6$. Similarly, if $0 \in{c, d}$, then $0 \in{a, b}$ and this contradicts $a b^{2}+c d^{2}=-6$ again. Hence $a, b, c, d \neq 0$.

Let the four equations in the problem be (i), (ii), (iii) and (iv), respectively. Then $(i)+3(i i)+3(i i i)+(i v)$ will give

$$(a+b{)}^3+(c+d{)}^3=-15.(a+b)^\{3\}+(c+d)^\{3\}=-15\; .$$

According to the equation (ii), $b$ and $d$ have different sign, and similarly (iv) yields that $a$ and $c$ have different sign.

First, consider the case $a>0, b>0$. Then $c<0$ and $d<0$. By $(i)$, we have $a>-c$ (i.e. $|a|>|c|)$ and (iii) gives $b>-d$. Hence $a+b>-(c+d)$ and so $(a+b)^{3}>-(c+d)^{3}$, thus $(a+b)^{3}+(c+d)^{3}>0$ which contradicts (1).

Next, consider the case $a>0, b<0$. Then $c<0$ and $d>0$. By $(i)$, we have $a>-c$ and by (iii), $d>-b$ (i.e. $b>-d$ ). Thus $a+b>-(c+d)$ and hence $(a+b)^{3}+(c+d)^{3}>0$ which contradicts (1).

The case $a<0, b<0$ leads to $c>0, d>0$. By (i), we have $c>-a$ and by (iii) $d>-b$. So $c+d>-(a+b)$ and hence $(c+d)^{3}+(a+b)^{3}>0$ which contradicts (1) again.

Finally, consider the case $a<0, b>0$. Then $c>0$ and $d<0$. By (i), $c>-a$ and by (iii) $b>-d$ which gives $c+d>-(a+b)$ and hence $(c+d)^{3}+(a+b)^{3}>0$ contradicting (1).

Hence there is no real solution to this system of equations. (Heiki Niglas)

Solution 3. As in Solution 2, we conclude that $a, b, c, d \neq 0$. The equation $a^{2} b+c^{2} d=0$ yields $a= \pm \sqrt{\frac{-d}{b}} c$. On the other hand, we have $a^{3}+c^{3}=2$ and $a b^{2}+c d^{2}=-6<0$ which implies that $\min {a, c}<0<\max {a, c}$ and thus $a=-\sqrt{\frac{-d}{b}} c$.

Let $x=-\sqrt{\frac{-d}{b}}$. Then $a=x c$ and so

$$2=a^3+c^3=c^3(1+x^3)\mathrm{.}2=a^\{3\}+c^\{3\}=c^\{3\}\backslash left(1+x^\{3\}\backslash right)\; .$$

Also $-6=a b^{2}+c d^{2}=c x b^{2}+c d^{2}$, which, using (2), gives

$${(x{b}^2+d^2)}^3=\frac{-6^3}{c^3}=-108(x^3+1)\text{. }\backslash left(x\; b^\{2\}+d^\{2\}\backslash right)^\{3\}=\backslash frac\{-6^\{3\}\}\{c^\{3\}\}=-108\backslash left(x^\{3\}+1\backslash right)\; \backslash text\; \{.\; \}$$

Thus

$$-108(1+x^3)={\left(d^2(x\frac{b^2}{d^2}+1)\right)}^3=d^6{(\frac{1}{x^3}+1)}^3=d^6{\left(\frac{1+x^3}{x^3}\right)}^3\mathrm{.}-108\backslash left(1+x^\{3\}\backslash right)=\backslash left(d^\{2\}\backslash left(x\; \backslash frac\{b^\{2\}\}\{d^\{2\}\}+1\backslash right)\backslash right)^\{3\}=d^\{6\}\backslash left(\backslash frac\{1\}\{x^\{3\}\}+1\backslash right)^\{3\}=d^\{6\}\backslash left(\backslash frac\{1+x^\{3\}\}\{x^\{3\}\}\backslash right)^\{3\}\; .$$

If $x^{3}+1=0$, then $x=-1$ and hence $a=-c$, which contradicts $a^{3}+c^{3}=2$. So $x^{3}+1 \neq 0$ and (3) gives

$$d^6{(1+x^3)}^2=-108x^9\mathrm{.}d^\{6\}\backslash left(1+x^\{3\}\backslash right)^\{2\}=-108\; x^\{9\}\; .$$

Now note that

$$x^3={(-\sqrt{\frac{-d}{b}})}^3=-\sqrt{\frac{-d^3}{b^3}}=-\sqrt{\frac{b^3-1}{b^3}}x^\{3\}=\backslash left(-\backslash sqrt\{\backslash frac\{-d\}\{b\}\}\backslash right)^\{3\}=-\backslash sqrt\{\backslash frac\{-d^\{3\}\}\{b^\{3\}\}\}=-\backslash sqrt\{\backslash frac\{b^\{3\}-1\}\{b^\{3\}\}\}$$

and hence (4) yields that

$${(b^3-1)}^2{(1-\sqrt{\frac{b^3-1}{b^3}})}^2=108{\left(\sqrt{\frac{b^3-1}{b^3}}\right)}^3\mathrm{.}\backslash left(b^\{3\}-1\backslash right)^\{2\}\backslash left(1-\backslash sqrt\{\backslash frac\{b^\{3\}-1\}\{b^\{3\}\}\}\backslash right)^\{2\}=108\backslash left(\backslash sqrt\{\backslash frac\{b^\{3\}-1\}\{b^\{3\}\}\}\backslash right)^\{3\}\; .$$

Let $y=\sqrt{\frac{b^{3}-1}{b^{3}}}$. Then $b^{3}=\frac{1}{1-y^{2}}$ and so (5) implies

$${(\frac{1}{1-y^2}-1)}^2{(1-y^2)}^2=108y^3\backslash left(\backslash frac\{1\}\{1-y^\{2\}\}-1\backslash right)^\{2\}\backslash left(1-y^\{2\}\backslash right)^\{2\}=108\; y^\{3\}$$

i.e.

$$\frac{y^4}{{(1-y^2)}^2}(1-y{)}^2=108y^3\backslash frac\{y^\{4\}\}\{\backslash left(1-y^\{2\}\backslash right)^\{2\}\}(1-y)^\{2\}=108\; y^\{3\}$$

If $y=0$, then $b=1$ and so $d=0$, a contradiction. So

$$y(1-y{)}^2=108(1-y{)}^2(1+y{)}^{2}y(1-y)^\{2\}=108(1-y)^\{2\}(1+y)^\{2\}$$

Clearly $y \neq 1$ and hence $y=108+108 y^{2}+216 y$, or $108 y^{2}+215 y+108=0$. The last equation has no real solutions and thus the initial system of equations has no real solutions.

Remark 1. Note that this solution worked because RHS of $a^{2} b+c^{2} d=0$ is zero. If instead it was, e.g., $a^{2} b+c^{2} d=0.1$ then this solution would not work out, but the first solution still would.

Remark 2. The advantage of this solution is that solving the last equation $108 y^{2}+$ $215 y+108=0$ one can find complex solutions of this system of equations. (Heiki Niglas)

10. Let $a_{0,1}, a_{0,2}, \ldots, a_{0,2016}$ be positive real numbers. For $n \geq 0$ and $1 \leq k<2016$ set

$$a_{n+1,k}=a_{n,k}+\frac{1}{2a_{n,k+1}}\text{ and }{a}_{n+1,2016}=a_{n,2016}+\frac{1}{2a_{n,1}}\mathrm{.}a\_\{n+1,\; k\}=a\_\{n,\; k\}+\backslash frac\{1\}\{2\; a\_\{n,\; k+1\}\}\; \backslash quad\; \backslash text\; \{\; and\; \}\; \backslash quad\; a\_\{n+1,2016\}=a\_\{n,\; 2016\}+\backslash frac\{1\}\{2\; a\_\{n,\; 1\}\}\; .$$

Show that $\max {1 \leq k \leq 2016} a{2016, k}>44$.

Solution. We prove

$$m_n^2\ge nm\_\{n\}^\{2\}\; \backslash geq\; n$$

for all $n$. The claim then follows from $44^{2}=1936<2016$. To prove (1), first notice that the inequality certainly holds for $n=0$. Assume (1) is true for $n$. There is a $k$ such that $a_{n, k}=m_{n}$. Also $a_{n, k+1} \leq m_{n}$ (or if $k=2016, a_{n, 1} \leq m_{n}$ ). Now (assuming $k<2016$ )

$$a_{n+1,k}^2={(m_n+\frac{1}{2a_{n,k+1}})}^2=m_n^2+\frac{m_n}{a_{n,k+1}}+\frac{1}{4a_{n,k+1}^2}>n+1.a\_\{n+1,\; k\}^\{2\}=\backslash left(m\_\{n\}+\backslash frac\{1\}\{2\; a\_\{n,\; k+1\}\}\backslash right)^\{2\}=m\_\{n\}^\{2\}+\backslash frac\{m\_\{n\}\}\{a\_\{n,\; k+1\}\}+\backslash frac\{1\}\{4\; a\_\{n,\; k+1\}^\{2\}\}>n+1\; .$$

Since $m_{n+1}^{2} \geq a_{n+1, k}^{2}$, we are done.

11. The set $A$ consists of 2016 positive integers. All prime divisors of these numbers are smaller than 30. Prove that there are four distinct numbers $a, b, c$ and $d$ in $A$ such that abcd is a perfect square.

Solution. There are ten prime numbers $\leq 29$. Let us denote them as $p_{1}, p_{2}, \ldots, p_{10}$. To each number $n$ in $A$ we can assign a 10 -element sequence $\left(n_{1}, n_{2}, \ldots, n_{10}\right)$ such that $n_{i}=1$ $p_{i}$ has an odd exponent in the prime factorization of $n$, and $n_{i}=0$ otherwise. Two numbers to which identical sequences are assigned, multiply to a perfect square. There are only 1024 different 10-element ${0,1}$-sequences so there exist some two numbers $a$ and $b$ with identical sequencies, and after removing these from $A$ certainly two other numbers $c$ and $d$ with identical sequencies remain. These $a, b, c$ and $d$ satisfy the condition of the problem.

12. Does there exist a hexagon (not necessarily convex) with side lengths 1, 2, 3, 4, 5, 6 (not necessarily in this order) that can be tiled with a) 31 b) 32 equilateral triangles with side length 1 ?

Solution. The adjoining figure shows that question a) can be answered positively.

For a negative answer to b), we show that the number of triangles has to be odd. Assume there are $x$ triangles in the triangulation. They hav altogether $3 x$ sides. Of these, $1+2+3+4+6=21$ are on the perimeter of the hexagon. The remaining $3 x-21$ sides are in the interior, and they touch each other pairwise. So $3 x-21$ has to be even, which is only possible, if $x$ is odd.

13. Let $n$ numbers all equal to 1 be written on a blackboard. A move consists of replacing two numbers on

the board with two copies of their sum. It happens that after $h$ moves all $n$ numbers on the blackboard are equal to $m$. Prove that $h \leq \frac{1}{2} n \log _{2} m$.

Solution. Let the product of the numbers after the $k$-th move be $a_{k}$. Suppose the numbers involved in a move were $a$ and $b$. By the arithmetic-geometric mean inequality, $(a+b)(a+b) \geq 4 a b$. Therefore, regardless of the choice of the numbers in the move, $a_{k} \geq 4 a_{k-1}$, and since $a_{0}=1, a_{h}=m^{n}$, we have $m^{n} \geq 4^{h}=2^{2 h}$ and $h \leq \frac{1}{2} n \log _{2} m$.

14. A cube consists of $4^{3}$ unit cubes each containing an integer. At each move, you choose a unit cube and increase by 1 all the integers in the neighbouring cubes having a face in common with the chosen cube. Is it possible to reach a position where all the $4^{3}$ integers are divisible by 3 , no matter what the starting position is?

Solution. Two unit cubes with a common face are called neighbours. Colour the cubes either black or white in such a way that two neighbours always have different colours. Notice that the integers in the white cubes only change when a black cube is chosen. Now recolour the white cubes that have exactly 4 neighbours and make them green. If we look at a random black cube it has either 0,3 or 6 white neighbours. Hence if we look at the sum of the integers in the white cubes, it changes by 0,3 or 6 in each turn. From this it follows that if this sum is not divisible by 3 at the beginning, it will never be, and none of the integers in the white cubes is divisible by 3 at any state.

15. The Baltic Sea has 2016 harbours. There are two-way ferry connections between some of them. It is impossible to make a sequence of direct voyages $C_{1}-C_{2}-\cdots-C_{1062}$ where all the harbours $C_{1}, \ldots, C_{1062}$ are distinct. Prove that there exist two disjoint sets $A$ and $B$ of 477 harbours each, such that there is no harbour in $A$ with a direct ferry connection to a harbour in $B$.

Solution. Let $V$ be the set of all harbours. Take any harbour $C_{1}$ and set $U=V \backslash\left{C_{1}\right}$, $W=\emptyset$. If there is a ferry connection from $C$ to another harbour, say $C_{2}$ in $V$, consider the route $C_{1} C_{2}$ and remove $C_{2}$ from $U$. Extend it as long as possible. Since there is no route of length 1061, So we have a route from $C_{1}$ to some $C_{k}, k \leq 1061$, and no connection from $C_{k}$ to a harbor not already included in the route exists. There are at least $2016-1062$ harbours in $U$. Now we move $C_{k}$ from $U$ to $W$ and try to extend the route from $C_{k-1}$ onwards. The extension again terminates at some harbor, which we then move from $U$ to $W$. If no connection from $C_{1}$ to any harbour exists, we move $C_{1}$ to $W$ and start the process again from some other harbour. This algorithm produces two sets of harbours, $W$ and $U$, between which there are no direct connections. During the process, the number of harbours in $U$ always decreases by 1 and the number of harbours in $W$ increases by 1 . So at some point the number of harbours is the same, and it then is at least $\frac{1}{2}(2016-1062)=477$. By removing, if necessary, some harbours fron $U$ and $W$ we get sets of exactly 477 harbours.

16. In triangle $A B C$, the points $D$ and $E$ are the intersections of the angular bisectors from $C$ and $B$ with the sides $A B$ and $A C$, respectively. Points $F$ and $G$ on the extensions of $A B$ and $A C$ beyond $B$ and $C$, respectively, satisfy $B F=C G=B C$. Prove that $F G | D E$.

Solution. Since $B E$ and $C D$ are angle bisectors,

$$\frac{AD}{AB}=\frac{AC}{AC+BC},\frac{AE}{AC}=\frac{AB}{AB+BC}\mathrm{.}\backslash frac\{A\; D\}\{A\; B\}=\backslash frac\{A\; C\}\{A\; C+B\; C\},\; \backslash quad\; \backslash frac\{A\; E\}\{A\; C\}=\backslash frac\{A\; B\}\{A\; B+B\; C\}\; .$$

So

$$\frac{AD}{AF}=\frac{AD}{AB}\cdot \frac{AB}{AF}=\frac{AC\cdot AB}{(AC+BC)(AB+BC)}\backslash frac\{A\; D\}\{A\; F\}=\backslash frac\{A\; D\}\{A\; B\}\; \backslash cdot\; \backslash frac\{A\; B\}\{A\; F\}=\backslash frac\{A\; C\; \backslash cdot\; A\; B\}\{(A\; C+B\; C)(A\; B+B\; C)\}$$

and

$$\frac{AE}{AG}=\frac{AE}{AC}\cdot \frac{AC}{AG}=\frac{AB\cdot AC}{(AB+AC)(AC+BC)}\backslash frac\{A\; E\}\{A\; G\}=\backslash frac\{A\; E\}\{A\; C\}\; \backslash cdot\; \backslash frac\{A\; C\}\{A\; G\}=\backslash frac\{A\; B\; \backslash cdot\; A\; C\}\{(A\; B+A\; C)(A\; C+B\; C)\}$$

Since $\frac{A D}{A F}=\frac{A E}{A G}, D E$ and $F G$ are parallel.

17. Let $A B C D$ be a convex quadrilateral with $A B=A D$. Let $T$ be a point on the diagonal $A C$ such that $\angle A B T+\angle A D T=\angle B C D$. Prove that $A T+A C \geq A B+A D$.

Solution. On the segment $A C$, consider the unique point $T^{\prime}$ such that $A T^{\prime} \cdot A C=A B^{2}$. The triangles $A B C$ and $A T^{\prime} B$ are similar: they have the angle at $A$ common, and $A T^{\prime}: A B=A B: A C$. So $\angle A B T^{\prime}=\angle A C B$. Analogously, $\angle A D T^{\prime}=\angle A C D$. So $\angle A B T^{\prime}+\angle A D T^{\prime}=\angle B C D$. But $A B T^{\prime}+A D T^{\prime}$ increases strictly monotonously, as $T^{\prime}$ moves from $A$ towards $C$ on $A C$. The assumption on $T$ implies that $T^{\prime}=T$. So, by the arithmetic-geometric mean inequality,

$$AB+AD=2AB=2\sqrt{AT\cdot AC}\le AT+AC\mathrm{.}A\; B+A\; D=2\; A\; B=2\; \backslash sqrt\{A\; T\; \backslash cdot\; A\; C\}\; \backslash leq\; A\; T+A\; C\; .$$

18. Let $A B C D$ be a parallelogram such that $\angle B A D=60^{\circ}$. Let $K$ and $L$ be the midpoints of $B C$ and $C D$, respectively. Assuming that $A B K L$ is a cyclic quadrilateral, find $\angle A B D$.

Solution. Let $\angle B A L=\alpha$. Since $A B K L$ is cyclic, $\angle K K C=\alpha$. Because $L K | D B$ and $A B | D C$, we further have $\angle D B C=\alpha$ and $\angle A D B=\alpha$. Let $B D$ and $A L$ intersect at $P$. The triangles $A B P$ and $D B A$ have two equal angles, and hence $A B P \sim D B A$. So

$$\frac{AB}{DB}=\frac{BP}{AB}\backslash frac\{A\; B\}\{D\; B\}=\backslash frac\{B\; P\}\{A\; B\}$$

The triangles $A B P$ and $L D P$ are clearly similar with similarity ratio $2: 1$. Hence $B P=$ $\frac{2}{3} D B$. Inserting this into (1) we get

$$AB=\sqrt{\frac{2}{3}}\cdot DBA\; B=\backslash sqrt\{\backslash frac\{2\}\{3\}\}\; \backslash cdot\; D\; B$$

The sine theorem applied to $A B D$ (recall that $\angle D A B=60^{\circ}$ ) immediately gives

$$\sin \alpha =\frac{AB}{BD}\sin {60}^{\circ }=\sqrt{\frac{2}{3}}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{2}=\sin {45}^{\circ }\mathrm{.}\backslash sin\; \backslash alpha=\backslash frac\{A\; B\}\{B\; D\}\; \backslash sin\; 60^\{\backslash circ\}=\backslash sqrt\{\backslash frac\{2\}\{3\}\}\; \backslash cdot\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}=\backslash frac\{\backslash sqrt\{2\}\}\{2\}=\backslash sin\; 45^\{\backslash circ\}\; .$$

So $\angle A B D=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$.

19. Consider triangles in the plane where each vertex has integer coordinates. Such a triangle can be legally transformed by moving one vertex parallel to the opposite side to a different point with integer coordinates. Show that if two triangles have the same area, then there exists a series of legal transformations that transforms one to the other.

Solution. We will first show that any such triangle can be transformed to a special triangle whose vertices are at $(0,0),(0,1)$ and $(n, 0)$. Since every transformation preserves the triangle's area, triangles with the same area will have the same value for $n$.

Define th $y$-span of a triangle to be the difference between the largest and the smallest $y$ coordinate of its vertices. First we show that a triangle with a $y$-span greater than one can be transformed to a triangle with a strictly lower $y$-span.

Assume $A$ has the highest and $C$ the lowest $y$ coordinate of $A B C$. Shifting $C$ to $C^{\prime}$ by the vector $\overrightarrow{B A}$ results in the new triangle $A B C^{\prime}$ where $C^{\prime}$ has larger $y$ coordinate than $C$ baut lower than $A$, and $C^{\prime}$ has integer coordinates. If $A C$ is parallel to the $x$-axis, a horizontal shift of $B$ can be made to transform $A B C$ into $A B^{\prime} C$ where $B^{\prime} C$ is vertical, and then $A$ can be vertically shifted so that the $y$ coordinate of $A$ is between those of $B^{\prime}$ and $C$. Then the $y$-span of $A B^{\prime} C$ can be reduced in the manner described above. Continuing the process, one necessarily arrives at a triangle with $y$-span equal to 1 . Such a triangle then necessarily has one side, say $A C$, horizontal. A legal horizontal move can take $B$ to the a position $B^{\prime}$ where $A B^{\prime}$ is horizontal and $C$ has the highest $x$-coordinate. If $B^{\prime}$ is above $A C$, perform a vertical and a horizontal legal move to take $B^{\prime}$ to the origin; the result is a special triangle. If $B^{\prime}$ is below $A C$, legal transformation again can bring $B^{\prime}$ to the origin, and a final horizontal transformation of one vertex produces the desired special triangle.

The inverse of a legal transformation is again a legal transformation. Hence any two triangles having vertices with integer coordinates and same area can be legally transformed into each other via a special triangle.

20. Let $A B C D$ be a cyclic quadrilateral with $A B$ and $C D$ not parallel. Let $M$ be the midpoint of $C D$. Let $P$ be a point inside $A B C D$ such that $P A=P B=C M$. Prove that $A B, C D$ and the perpendicular bisector of $M P$ are concurrent.

Solution. Let $\omega$ be the circumcircle of $A B C D$. Let $A B$ and $C D$ intersect at $X$. Let $\omega_{1}$ and $\omega_{2}$ be the circles with centers $P$ and $M$ and with equal radius $P B=M C=r$. The power of $X$ with respect to $\omega$ and $\omega_{1}$ equals $X A \cdot X B$ and with respect to $\omega$ and $\omega_{2} X D \cdot X C$. The latter power also equals $(X M+$ $r)(X M-r)=X M^{2}-r^{2}$. Analogously, the first power is $X P^{2}-r^{2}$. But since $X A \cdot X B=X D \cdot X C$, we must have $X M^{2}=X P^{2}$ or $X M=X P . X$ indeed is on the perpendicular bisector of $P M$, and we are done.