42nd Balkan Mathematical Olympiad

If $k=1$, then $n_1!=a^{n_1}$. Since $n_1>1$, then $2|n_1!$ and so $2|a^{n_1}$ and therefore $2|a$. Then $$n_1\le v_2(a^{n_1})=v_2(n_1!)=\lfloor \frac{n_1}{2}\rfloor +\lfloor \frac{n_1}{2^2}\rfloor +\cdots <\frac{n_1}{2}+\frac{n_1}{2^2}+\cdots =n_1,$$ a contradiction. So we can assume $k\ge 2$. Let $m_1<m_2$ be the two smallest elements of $\{n_1,n_2,\dots ,n_k\}$ and assume for the sake of contradiction that they are both composite.

Case 1. Assume $m_2\ge m_1+2$. Then $v_2(n_i!)>v_2(m_1!)$ for each $i$ with $n_i\ne m_1$. Therefore $$n_1\le v_2(a^{n_1})=v_2(n_1!+n_2!+\cdots +n_k!)=v_2(m_1!)\le v_2(n_1!)<n_1,$$ a contradiction.

Case 2. Assume $m_2=m_1+1$. Let $p$ be a prime factor of $m_2$. Since $m_2$ is composite, then $p\le m_1$ and so $p|n_i!$ for each $i$, hence $p|a$. Furthermore, $v_p(n_i!)\ge v_p(m_2!)>v_p(m_1!)$ for each $i$ with $n_i\ne m_1$ (since $p|m+1$) and therefore $$n_1\le v_p(a^{n_1})=v_p(n_1!+n_2!+\cdots +n_k!)=v_p(m_1!)\le v_p(n_1!)=\lfloor \frac{n_1}{p}\rfloor +\lfloor \frac{n_1}{p^2}\rfloor +\cdots <n_1,$$ a contradiction.

Hence, we conclude that at least one of $m_1$ and $m_2$ has to be prime, which finishes the problem.

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Alternative solution.

We solve the case $k=1$ in the same way as in the first solution. Now, let $k\ge 2$ and let $n_s$ be the smallest element of $\{n_1,n_2,\dots ,n_k\}$. Assume for the sake of contradiction that all $n_1,n_2,\dots ,n_k$ are composite. Write the given equation in the following way: $$n_s!\cdot (1+(n_s+1)\dots n_1+\cdots +(n_s+1)\dots n_{s-1}+(n_s+1)\dots n_{s+1}+\cdots +(n_s+1)\dots n_k)=a^{n_1}.$$ Denote the second factor on the left hand side with $A$, so we have that $n_s!\cdot A=a^{n_1}$. Assume some prime number $p\le n_s$ doesn't divide $A$. We then have that $p|n_s!$, hence $p|a$ and: $$v_p(n_s!\cdot A)=v_p(n_s!)<n_s\le n_1\le v_p(a^{n_1})=v_p(n_s!\cdot A),$$ contradiction. Hence, all primes $p$ that divide $n_s!$ also divide $A$. Next, notice that if $n_s+1$ is composite, we can take a prime factor $q|n_s+1$ such that $q\le n_s$ (so $q|n_s!$), which implies that $q|A-1$ (since $A-1$ is divisible by $n_s+1$) which then implies that $q$ does not divide $A$, contradiction. Thus, $n_s+1=q$ for some prime number $q$. This means that $n_i\ge q+1=n_s+2$ for all $i\ne s$ (since by assumption none of the $n_i$ are prime). But, this also leads to a contradiction, since now we can pick some prime factor $r\le n_s$ of $n_s+2$ (note that $n_s+2=q+1$ can't be prime, since that would imply $q=2$ and $n_s=q-1=1$) and have that $r|A-1$ (since now $A-1$ is divisible by $n_s+2$), so $r$ does not divide $A$. Hence, we conclude that at least one of $n_1,n_2,\dots ,n_k$ has to be prime, which finishes the problem.