Consider the map $\phi$ whose domain is the set of positive integers with all prime factors less than $p_n$ and whose range is the vector space $\mathbf{Q}^{n-1}$ defined by $\phi(2^{a_1}\cdot 3^{a_2}\cdots p_{n-1}^{a_{n-1}})=(a_1,a_2,\ldots,a_{n-1})$. By applying $\phi$ everywhere, we may rephrase the problem in the following more general form: given a set of nonzero linearly dependent vectors in $\mathbf{Q}_{\geq0}^{m}$, is it possible by successively adding one vector to another (possibly the same one) to eventually produce two equal vectors? In this rephrasing, it is straightforward to show that the answer is affirmative: take a nontrivial linear dependence relation among the vectors. Since we're working over $\mathbf{Q}$ we may clear denominators in the coefficients to make this a dependence relation with integer coefficients. Since the vectors are nonzero with nonnegative entries, the dependence relation can be written in the form $c_1 v_1+\ldots=c'_1v'_1+\ldots$ where the $v$s and $v'$s are (some of) our vectors, the $c$s and $c'$s are positive integer coefficients, and neither side of the equation is an empty sum. The coefficients tell us exactly how to build two equal vectors.

JBL, I don't quite understand how you went from the dependence relation to a valid construction. For example, if we had the set of vectors \[ \begin{cases} v_1=(2,0,0,0) \\ v_2=(3,0,0,0) \\ v_3=(1,1,0,0) \\ v_4=(0,0,1,0) \\ v_5=(0,2,0,0) \end{cases} \] then we have $ 6v_3=2v_2+3v_5 $, but we can't add $ v_3 $ to itself six times, or $ v_5 $ to itself three times.

Ach, thanks, you're absolutely right -- we can only obviously get powers of 2. I'll have to think more about whether this approach can be repaired.

http://emc.mnm.hr/wp-content/uploads/2012/12/1st-EMC-Senior-Category-Solutions.pdf Well... but I can't understand the solution...

This problem is very hard and the official solution is slightly unclear. Here's a clearer write-up: For simplicity, let Olja be player A and Oleg be player B. The answer is yes for $n \geqslant 2$ and no for $n = 1.$ For $n = 1$, player B won’t be able to write 2 equal numbers as there is only one number written on the board. We shall now consider the case $n \geqslant 2.$ Note that because $a_1, a_2,\ldots a_n$ are all sirictly smaller than $p_n,$ their prime divisors belong to the set $\{p_1, p_2, \ldots, p_{n-1}\}.$ For each $i=1,2,\ldots,n$ let $\nu_{p_j}(a_i)=\alpha_{i,j}$ and consider the associated vector \[\vec{v}_i=\left(\alpha_{i,1},\alpha_{i,2},\ldots,\alpha_{i,n-1}\right).\]Note that the vectors $\vec{v}_1,\vec{v}_2,\ldots,\vec{v}_n$ are linearly dependent, that is, there exist rational numbers ${}x_1,x_2,\ldots,x_n$ not all of which are 0 for which $x_1\cdot\vec{v}_1+x_2\cdot\vec{v}_2+\cdots+x_n\cdot\vec{v}_n=\vec{0}.$ Therefore, \[\prod_{i=1}^na_i^{x_i}=\prod_{i=1}^{n}\prod_{j=1}^{n-1}p_j^{\alpha_{i,j}\cdot x_i}=\prod_{j=1}^{n-1}p_j^0=1.\]Write $x_1,x_2,\ldots,x_n{}$ in lowest terms and let $L{}$ be the greatest common divisor of their denominators. For $i=1,2,\ldots,n$ let $r_i=L\cdot x_i$ and note that $r_1,r_2,\ldots,r_n$ are relatively prime integers which satisfy $a_1^{r_1}\cdot a_2^{r_2}\cdots a_n^{r_n}=1.$ Now, let $1,\ldots,k$ be the indices for which $r_i<0,$ $k+1,\ldots,k+l$ be the indices for which $r_i>0,$ and $k+l+1,\ldots,n$ be the indices for which $r_i=0.$ Of course, $k,l\geqslant 1.$ Therefore, we have \[\prod_{i=1}^ka_i^{r_i}=\prod_{i=k+1}^{k+l}a_i^{-r_i},\]where $r_1,\ldots,r_k$ and $-r_{k+1},\ldots,-r_{k+l}$ are relatively prime positive integers. Lemma 1. Let $x_1,x_2$ be relatively prime positive integers. For any positive integers $a,b$ there exists a sequence of operations which replaces the numbers $(a,b)$ with the numbers $(a',b'),$ one of which is $a^{x_1}\cdot b^{x_2}.$ Proof. We will prove this by strong induction on $x_1+x_2,$ for all $a,b.$ The base case $x_1+x_2=2$ is trivial. Now, assume that the lemma holds for all $a,b$ when $x_1+x_2<n$ with $n\geqslant 3.$ We will prove that it also holds for $x_1+x_2=n.$ Note that since $\gcd(x_1,x_2)=1$ and $x_1+x_2>2$ then $x_1\neq x_2.$ Let $x_1>x_2$ and perform the operation $(a,b)\mapsto(a,ab)$ and then apply the inductive hypothesis on $(x_1-x_2,x_2)$ and $(a,ab)$ to finish the proof. $\square$ Lemma 2. Let $b_1,\ldots,b_k$ and $x_1,\ldots,x_k$ be $2k$ positive integers. Then, there exists a sequence of operations which replaces the numbers $(b_1,\ldots,b_k)$ with the numbers $(b_1',\ldots,b_k'),$ one of which is $b_1^{x_1/d}\cdots b_k^{x_k/d}$ where $d=\gcd(x_1,\ldots,x_k).$ Proof. We will prove this by induction on $k,$ for any $b_1,\ldots,b_k$ and $x_1,\ldots,x_k,$ the base case $k=1$ being trivial. Assume the lemma is true for some $k{}.$ We will prove that it also holds for $k+1.{}$ Take $b_1,\ldots,b_{k+1}$ and $x_1,\ldots,x_{k+1}.$ Apply Lemma 1 on $(b_k,b_{k+1})$ and $(x_k',x_{k+1}')$ where $x_k'=x_k/\gcd(x_k,x_{k+1})$ and $x_{k+1}'$ is defined analogously. Hence, there exists a sequence of operations which results in the transformation \[(b_1,\ldots,b_{k+1})\mapsto\left(b_1,\ldots,b_{k-1},\gamma,b_k^{x_k'}\cdot b_{k+1}^{x_k'}\right)\]where $\gamma$ is some positive integer. Apply the inductive hypothesis on $b_1,\ldots,b_{k-1},b_k^{x_k'}b_{k+1}^{x_k'}$ and $x_1,\ldots,x_{k-1},\gcd(x_k,x_{k+1})$ to finish the proof, making use of the fact that $\gcd(x_1,\ldots,x_{k-1},\gcd(x_k,x_{k+1}))=\gcd(x_1,\ldots,x_{k+1}).$ $\square$ Lemma 3. Let $x_1,x_2$ be relatively prime positive integers. For any positive integers $a,b$ there exists a sequence of operations which replaces the numbers $(a,b)$ with the numbers $(a',b'),$ which satisfy $a'/b'=a^{x_1}/b^{x_2}.$ Proof. We will prove this by strong induction on $x_1+x_2,$ for all $a,b.$ The base case $x_1+x_2=2$ is trivial. Now, assume that the lemma holds for all $a,b$ when $x_1+x_2<n$ with $n\geqslant 3.$ We will prove that it also holds for $x_1+x_2=n.$ Two cases arise. First, if one of $x_1,x_2$ is even, WLOG say $x_1,$ then make the operation $(a,b)\mapsto(a^2,b)$ and apply the inductive hypothesis for $(a^2,b)$ and $(x_1/2,x_2)$ and we're done. Clearly, $x_1,x_2$ must be distinct and cannot both be even. Hence, the second case is $x_1,x_2$ odd and distinct. Let $x_1>x_2.$ Perform the operations $(a,b)\mapsto(a,ab)\mapsto(a^2,ab)$ and apply the induction hypothesis on $(a^2,ab)$ and $((x_1+x_2)/2,x_2)$ to finish the proof. $\square$ We may now finally finish the solution. Let $d_1=\gcd(r_1,\ldots,r_k)$ and $d_2=\gcd(-r_{k+1},\ldots,-r_{k+l}).$ For $i=1,\ldots,k$ let $z_i=r_i/d_1$ and for $i=k+1,\ldots,k+l$ let $z_i=r_i/d_2.$ Furthermore, let $P{}$ denote the product on the left-hand side and $Q{}$ the one on the right-hand side and let $P'=P^{1/d_1}$ and $Q'=Q^{1/d_2}.$ Apply Lemma 2 on $a_1,\ldots,a_k$ and $z_1,\ldots,z_k$ in order to get $P'{}$ on the board and on $a_{k+1},\ldots,a_{k+l}$ and $z_{k+1},\ldots,z_{k+l}$ to get $Q'$ on the board (along with some other $k+l-2$ numbers we ignore). Note that since $r_1,\ldots,r_k,-r_{k+1},\ldots,-r_{k+l}$ are relatively prime, then $\gcd(d_1,d_2)=1$. Thus, we may apply Lemma 3 on $(P',Q')$ and $(d_1,d_2)$ to get two numbers $x,y$ on the board which satisfy $x/y=P'^{d_1}/Q'^{d_2}=P/Q=1$ as desired.