APMO/md/en-apmo2014_sol.md

# Solutions of APMO 2014 

Problem 1. For a positive integer $m$ denote by $S(m)$ and $P(m)$ the sum and product, respectively, of the digits of $m$. Show that for each positive integer $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the following conditions:

$$
S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right) \text { and } S\left(a_{i}\right)=P\left(a_{i+1}\right) \quad(i=1,2, \ldots, n)
$$

(We let $\left.a_{n+1}=a_{1}.\right)$ (Problem Committee of the Japan Mathematical Olympiad Foundation)
Solution. Let $k$ be a sufficiently large positive integer. Choose for each $i=2,3, \ldots, n$, $a_{i}$ to be a positive integer among whose digits the number 2 appears exactly $k+i-2$ times and the number 1 appears exactly $2^{k+i-1}-2(k+i-2)$ times, and nothing else. Then, we have $S\left(a_{i}\right)=2^{k+i-1}$ and $P\left(a_{i}\right)=2^{k+i-2}$ for each $i, 2 \leq i \leq n$. Then, we let $a_{1}$ be a positive integer among whose digits the number 2 appears exactly $k+n-1$ times and the number 1 appears exactly $2^{k}-2(k+n-1)$ times, and nothing else. Then, we see that $a_{1}$ satisfies $S\left(a_{1}\right)=2^{k}$ and $P\left(a_{1}\right)=2^{k+n-1}$. Such a choice of $a_{1}$ is possible if we take $k$ to be large enough to satisfy $2^{k}>2(k+n-1)$ and we see that the numbers $a_{1}, \ldots, a_{n}$ chosen this way satisfy the given requirements.

Problem 2. Let $S=\{1,2, \ldots, 2014\}$. For each non-empty subset $T \subseteq S$, one of its members is chosen as its representative. Find the number of ways to assign representatives to all non-empty subsets of $S$ so that if a subset $D \subseteq S$ is a disjoint union of non-empty subsets $A, B, C \subseteq S$, then the representative of $D$ is also the representative of at least one of $A, B, C$. (Warut Suksompong, Thailand)

Solution. Answer: 108 - 2014!.
For any subset $X$ let $r(X)$ denotes the representative of $X$. Suppose that $x_{1}=r(S)$. First, we prove the following fact:

$$
\text { If } x_{1} \in X \text { and } X \subseteq S \text {, then } x_{1}=r(X) \text {. }
$$

If $|X| \leq 2012$, then we can write $S$ as a disjoint union of $X$ and two other subsets of $S$, which gives that $x_{1}=r(X)$. If $|X|=2013$, then let $y \in X$ and $y \neq x_{1}$. We can write $X$ as a disjoint union of $\left\{x_{1}, y\right\}$ and two other subsets. We already proved that $r\left(\left\{x_{1}, y\right\}\right)=x_{1}$ (since $\left|\left\{x_{1}, y\right\}\right|=2<2012$ ) and it follows that $y \neq r(X)$ for every $y \in X$ except $x_{1}$. We have proved the fact.

Note that this fact is true and can be proved similarly, if the ground set $S$ would contain at least 5 elements.

There are 2014 ways to choose $x_{1}=r(S)$ and for $x_{1} \in X \subseteq S$ we have $r(X)=x_{1}$. Let $S_{1}=S \backslash\left\{x_{1}\right\}$. Analogously, we can state that there are 2013 ways to choose $x_{2}=r\left(S_{1}\right)$ and for $x_{2} \in X \subseteq S_{1}$ we have $r(X)=x_{2}$. Proceeding similarly (or by induction), there are $2014 \cdot 2013 \cdots 5$ ways to choose $x_{1}, x_{2}, \ldots, x_{2010} \in S$ so that for all $i=1,2 \ldots, 2010$, $x_{i}=r(X)$ for each $X \subseteq S \backslash\left\{x_{1}, \ldots, x_{i-1}\right\}$ and $x_{i} \in X$.

We are now left with four elements $Y=\left\{y_{1}, y_{2}, y_{3}, y_{4}\right\}$. There are 4 ways to choose $r(Y)$. Suppose that $y_{1}=r(Y)$. Then we clearly have $y_{1}=r\left(\left\{y_{1}, y_{2}\right\}\right)=r\left(\left\{y_{1}, y_{3}\right\}\right)=r\left(\left\{y_{1}, y_{4}\right\}\right)$. The only subsets whose representative has not been assigned yet are $\left\{y_{1}, y_{2}, y_{3}\right\},\left\{y_{1}, y_{2}, y_{4}\right\}$, $\left\{y_{1}, y_{3}, y_{4}\right\},\left\{y_{2}, y_{3}, y_{4}\right\},\left\{y_{2}, y_{3}\right\},\left\{y_{2}, y_{4}\right\},\left\{y_{3}, y_{4}\right\}$. These subsets can be assigned in any way, hence giving $3^{4} \cdot 2^{3}$ more choices.

In conclusion, the total number of assignments is $2014 \cdot 2013 \cdots 4 \cdot 3^{4} \cdot 2^{3}=108 \cdot 2014$ !.
Problem 3. Find all positive integers $n$ such that for any integer $k$ there exists an integer $a$ for which $a^{3}+a-k$ is divisible by $n$. (Warut Suksompong, Thailand)

Solution. Answer: All integers $n=3^{b}$, where $b$ is a nonnegative integer.
We are looking for integers $n$ such that the set $A=\left\{a^{3}+a \mid a \in \mathbf{Z}\right\}$ is a complete residue system by modulo $n$. Let us call this property by $\left(^{*}\right)$. It is not hard to see that $n=1$ satisfies $\left({ }^{*}\right)$ and $n=2$ does not.

If $a \equiv b(\bmod n)$, then $a^{3}+a \equiv b^{3}+b(\bmod n)$. So $n$ satisfies $\left(^{*}\right)$ iff there are no $a, b \in\{0, \ldots, n-1\}$ with $a \neq b$ and $a^{3}+a \equiv b^{3}+b(\bmod n)$.

First, let us prove that $3^{j}$ satisfies $\left(^{*}\right)$ for all $j \geq 1$. Suppose that $a^{3}+a \equiv b^{3}+b\left(\bmod 3^{j}\right)$ for $a \neq b$. Then $(a-b)\left(a^{2}+a b+b^{2}+1\right) \equiv 0\left(\bmod 3^{j}\right)$. We can easily check mod 3 that $a^{2}+a b+b^{2}+1$ is not divisible by 3 .

Next note that if $A$ is not a complete residue system modulo integer $r$, then it is also not a complete residue system modulo any multiple of $r$. Hence it remains to prove that any prime $p>3$ does not satisfy (*).

If $p \equiv 1(\bmod 4)$, there exists $b$ such that $b^{2} \equiv-1(\bmod p)$. We then take $a=0$ to obtain the congruence $a^{3}+a \equiv b^{3}+b(\bmod p)$.

Suppose now that $p \equiv 3(\bmod 4)$. We will prove that there are integers $a, b \not \equiv 0(\bmod p)$ such that $a^{2}+a b+b^{2} \equiv-1(\bmod p)$. Note that we may suppose that $a \not \equiv b(\bmod p)$, since otherwise if $a \equiv b(\bmod p)$ satisfies $a^{2}+a b+b^{2}+1 \equiv 0(\bmod p)$, then $(2 a)^{2}+(2 a)(-a)+$ $a^{2}+1 \equiv 0(\bmod p)$ and $2 a \not \equiv-a(\bmod p)$. Letting $c$ be the inverse of $b$ modulo $p$ (i.e. $b c \equiv 1(\bmod p)$ ), the relation is equivalent to $(a c)^{2}+a c+1 \equiv-c^{2}(\bmod p)$. Note that $-c^{2}$ can take on the values of all non-quadratic residues modulo $p$. If we can find an integer $x$ such that $x^{2}+x+1$ is a non-quadratic residue modulo $p$, the values of $a$ and $c$ will follow immediately. Hence we focus on this latter task.

Note that if $x, y \in\{0, \ldots, p-1\}=B$, then $x^{2}+x+1 \equiv y^{2}+y+1(\bmod p)$ iff $p$ divides $x+y+1$. We can deduce that $x^{2}+x+1$ takes on $(p+1) / 2$ values as $x$ varies in $B$. Since there are $(p-1) / 2$ non-quadratic residues modulo $p$, the $(p+1) / 2$ values that $x^{2}+x+1$ take on must be 0 and all the quadratic residues.

Let $C$ be the set of quadratic residues modulo $p$ and 0 , and let $y \in C$. Suppose that $y \equiv z^{2}(\bmod p)$ and let $z \equiv 2 w+1(\bmod p)$ (we can always choose such $\left.w\right)$. Then $y+3 \equiv$ $4\left(w^{2}+w+1\right)(\bmod p)$. From the previous paragraph, we know that $4\left(w^{2}+w+1\right) \in C$. This means that $y \in C \Longrightarrow y+3 \in C$. Unless $p=3$, the relation implies that all elements of $B$ are in $C$, a contradiction. This concludes the proof.

Problem 4. Let $n$ and $b$ be positive integers. We say $n$ is $b$-discerning if there exists a set consisting of $n$ different positive integers less than $b$ that has no two different subsets $U$ and $V$ such that the sum of all elements in $U$ equals the sum of all elements in $V$.
(a) Prove that 8 is a 100 -discerning.
(b) Prove that 9 is not 100-discerning.
(Senior Problems Committee of the Australian Mathematical Olympiad Committee)

## Solution.

(a) Take $S=\{3,6,12,24,48,95,96,97\}$, i.e.

$$
S=\left\{3 \cdot 2^{k}: 0 \leq k \leq 5\right\} \cup\left\{3 \cdot 2^{5}-1,3 \cdot 2^{5}+1\right\}
$$

As $k$ ranges between 0 to 5 , the sums obtained from the numbers $3 \cdot 2^{k}$ are $3 t$, where $1 \leq t \leq 63$. These are 63 numbers that are divisible by 3 and are at most $3 \cdot 63=189$.

Sums of elements of $S$ are also the numbers $95+97=192$ and all the numbers that are sums of 192 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all divisible by 3 and at least equal to 192. In addition, sums of elements of $S$ are the numbers 95 and all the numbers that are sums of 95 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $-1 \bmod$ 3.

Finally, sums of elements of $S$ are the numbers 97 and all the numbers that are sums of 97 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $1 \bmod 3$.

Hence there are at least $63+64+64+64=255$ different sums from elements of $S$. On the other hand, $S$ has $2^{8}-1=255$ non-empty subsets. Therefore $S$ has no two different subsets with equal sums of elements. Therefore, 8 is 100 -discerning.
(b) Suppose that 9 is 100 -discerning. Then there is a set $S=\left\{s_{1}, \ldots, s_{9}\right\}, s_{i}<100$ that has no two different subsets with equal sums of elements. Assume that $0<s_{1}<\cdots<s_{9}<$ 100.

Let $X$ be the set of all subsets of $S$ having at least 3 and at most 6 elements and let $Y$ be the set of all subsets of $S$ having exactly 2 or 3 or 4 elements greater than $s_{3}$.

The set $X$ consists of

$$
\binom{9}{3}+\binom{9}{4}+\binom{9}{5}+\binom{9}{6}=84+126+126+84=420
$$

subsets of $S$. The set in $X$ with the largest sums of elements is $\left\{s_{4}, \ldots, s_{9}\right\}$ and the smallest sums is in $\left\{s_{1}, s_{2}, s_{3}\right\}$. Thus the sum of the elements of each of the 420 sets in $X$ is at least $s_{1}+s_{2}+s_{3}$ and at most $s_{4}+\cdots+s_{9}$, which is one of $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1$ integers. From the pigeonhole principle it follows that $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1 \geq 420$, i.e.,

$$
\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right) \geq 419
$$

Now let us calculate the number of subsets in $Y$. Observe that $\left\{s_{4}, \ldots, s_{9}\right\}$ has $\binom{6}{2}$ 2-element subsets, $\binom{6}{3}$ 3-element subsets and $\binom{6}{4}$ 4-element subsets, while $\left\{s_{1}, s_{2}, s_{3}\right\}$ has exactly 8 subsets. Hence the number of subsets of $S$ in $Y$ equals

$$
8\left(\binom{6}{2}+\binom{6}{3}+\binom{6}{4}\right)=8(15+20+15)=400
$$

The set in $Y$ with the largest sum of elements is $\left\{s_{1}, s_{2}, s_{3}, s_{6}, s_{7}, s_{8}, s_{9}\right\}$ and the smallest sum is in $\left\{s_{4}, s_{5}\right\}$. Again, by the pigeonhole principle it follows that $\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+\right.$ $\left.s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right)+1 \geq 400$, i.e.,

$$
\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right) \geq 399
$$

Adding (1) and (2) yields $2\left(s_{6}+s_{7}+s_{8}+s_{9}\right) \geq 818$, so that $s_{9}+98+97+96 \geq$ $s_{9}+s_{8}+s_{7}+s_{6} \geq 409$, i.e. $s_{9} \geq 118$, a contradiction with $s_{9}<100$. Therefore, 9 is not 100-discerning.

Problem 5. Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the $\operatorname{arc} A B$ of circle $\omega$ ( $M$ lies inside $\Omega$ ). A chord $M P$ of circle $\omega$ intersects $\Omega$ at $Q(Q$ lies inside $\omega)$. Let $\ell_{P}$ be the tangent line to $\omega$ at $P$, and let $\ell_{Q}$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_{P}, \ell_{Q}$, and $A B$ is tangent to $\Omega$. (Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan)

Solution. Denote $X=A B \cap \ell_{P}, Y=A B \cap \ell_{Q}$, and $Z=\ell_{P} \cap \ell_{Q}$. Without loss of generality we have $A X<B X$. Let $F=M P \cap A B$.
![](https://cdn.mathpix.com/cropped/2024_11_22_fa403cf18214215ea344g-4.jpg?height=1169&width=1161&top_left_y=630&top_left_x=487)

Denote by $R$ the second point of intersection of $P Q$ and $\Omega$; by $S$ the point of $\Omega$ such that $S R \| A B$; and by $T$ the point of $\Omega$ such that $R T \| \ell_{P}$. Since $M$ is the midpoint of arc $A B$, the tangent $\ell_{M}$ at $M$ to $\omega$ is parallel to $A B$, so $\angle(A B, P M)=\angle\left(P M, \ell_{P}\right)$. Therefore we have $\angle P R T=\angle M P X=\angle P F X=\angle P R S$. Thus the point $Q$ is the midpoint of the $\operatorname{arc} T Q S$ of $\Omega$, hence $S T \| \ell_{Q}$. So the corresponding sides of the triangles $R S T$ and $X Y Z$ are parallel, and there exist a homothety $h$ mapping $R S T$ to $X Y Z$.

Let $D$ be the second point of intersection of $X R$ and $\Omega$. We claim that $D$ is the center of the homothety $h$; since $D \in \Omega$, this implies that the circumcircles of triangles $R S T$ and $X Y Z$ are tangent, as required. So, it remains to prove this claim. In order to do this, it suffices to show that $D \in S Y$.

By $\angle P F X=\angle X P F$ we have $X F^{2}=X P^{2}=X A \cdot X B=X D \cdot X R$. Therefore, $\frac{X F}{X D}=\frac{X R}{X F}$, so the triangles $X D F$ and $X F R$ are similar, hence $\angle D F X=\angle X R F=\angle D R Q=$ $\angle D Q Y$; thus the points $D, Y, Q$, and $F$ are concyclic. It follows that $\angle Y D Q=\angle Y F Q=$ $\angle S R Q=180^{\circ}-\angle S D Q$ which means exactly that the points $Y, D$, and $S$ are collinear, with $D$ between $S$ and $Y$.