I claim that only $n=7$ works and it's easy to verify. Let $p$ be the largest prime atmost $n$. Since $p\mid n!$, there exists two primes $p_1,p_2$, such that $p\mid p_1+p_2<2p$. Hence $p_1+p_2=p$. Since $p$ is odd, $p_1=2$. Call $p_2=q$ from now on. Note similarly, $q\mid n!$, so there are primes $r,r'$ such that. $q\mid r+r'< 3q$. We consider two cases: If $r+r'=q$, then as before we get $r'=2$ and $r=q-2$. If $r+r'=2q$, then one of them, say $r'$ must equal $p$ for size reasons, so we again get $r=q-2$. In both cases we see that $r,r+2,r+4$ are all primes, and one of them must be divisible by $3$. This proves $r+4=p=7$. One can verify that $n=8,9,10$ don't work. Done. $\square$

Aryan-23 wrote: I claim that only $n=7$ works and it's easy to verify. This is incorrect, since $n = 1$ also works (as the product is empty and we have $1! \mid 1$). The official solution I received actually didn't include $n = 1$, so presumably the PSC forgot to include an $n \ge 3$ assertion in order to make the product nonempty. A funny mistake to make, I think.

We show that $n<11$ from which $n=7$ is the only solution after checking cases. Assume $n\ge 11$. Suppose $r$ is the largest prime with $r\le n$ (so that $r\ge 11$). Then there are no two distinct primes $p,q\le n$ such that $p+q=2r$, so there must be two distinct primes $p,q\le n$ such that $p+q=r$ (as $r|n!$). By parity, we require $r-2$ to be prime as well. Now since $(r-2)|n!$ we need to find two primes $p,q\le n$ such that $(r-2)|p+q$. Thus $p+q=2(r-2)$ or $p+q=r-2$. To add up to $2(r-2)$, we need $r$ and $r-4$, but $r-4$ cannot be prime if $r,r-2$ are prime (primes are $\pm 1\pmod{6}$). To add up to $r-2$, by parity again we require $r-4$ prime, which is impossible.

The answer is only $n=7$. If $n<11$, we can easily check that this is the only solution (actually $n=1$ might work I'm not sure). Assume $n\geq 11$. Let $2\leq p_1 < p_2< \dots < p_k \leq n$ be the primes at most $n$. Since $p_k \mid \prod_{p < q\leq n}(p+q)$, we must have two primes $p_i, p_j$ such that $p_k\mid p_i + p_j$. Since $p_k$ is the largest prime, it is easy to see that $p_i +p_j=p_k$ and this implies $p_i = p_k-2$ and $i=k-1$, and $p_j=2$ in some order. Similarly, $p_{k-1} \mid \prod_{p<q \leq n} (p+q)$, and we must have $p_i+p_j=2p_{k-1}$ or $p_i + p_j = p_{k-1}$. Either case, we can easily determine that $p_{k-2} = p_{k-1}-2 = p_k-4$. Continuing downward, we see that all the odd numbers at most $n$ except $1$ are prime, but this is obviously false since $9$ is not prime. We are done.

Take the largest prime $p$ less than $n$. Now, the RHS must be $0 \pmod p$, and this is only possible if $p-2$ is prime because this forces the fact that two primes must sum to $p$. Now repeating this process for $p-2$, we get that $p-4$ must be prime. From here it's obvious via mod 3 or mod 5 and checking cases that the only answer is $n = 7$

Solved with plang2008. The answer is $\boxed{n = 1,7}$, which can be checked to work. Now we prove they are the only solutions. Assume henceforth that $n > 2$ because we can check that $n=2$ fails. Let $r$ be the largest prime less than or equal to $n$. We get that $r$ divides the RHS, so there exist distinct primes $a,b \le n$ so that $r$ divides $a + b$. However, $a, b < r$, so $ a+ b < 2r$, which implies $a + b = r$. Since $n > 2$, $r$ is odd, so either $a = 2$ or $b = 2$ (by parity). This implies $r - 2$ is prime. Now we know there exist distinct primes $c,d \le n$ such that $r-2\mid c + d$. Since $c,d$ are distinct, none of them can be equal to $r - 2$, so $c + d \le r + (r-4) = 2(r-2)$, which implies $c + d \in \{r-2, 2(r-2)\}$. Claim: $r-4$ is prime Proof: If $ c+ d = r-2$, then since $r-2$ is an odd prime, we can use parity and get that $c = 2$ or $d = 2$, so $r - 4 $ is prime. If $c + d = 2(r-2) = 2 r - 4$, then either $c = r$ or $d = r$, so $r -4 $ is prime. $\square$ Thus, $r, r-2, r-4$ are all prime, which is well known to imply $r = 7$. So $n \in \{7,8,9,10\}$. We note that the product on the RHS is equal to \[(2 + 3)(2 + 5)( 2+ 7)( 3 + 5)( 3 + 7)( 5 + 7) = 302400,\]which is a multiple of $7!$, but not $8!, 9!, $ or $10!$. This implies $n = 7$ works, but not $8,9,10$. Hence the only solution with $n > 1$ is $n = 7$.

The answer is $\boxed{n=1,7}$. We can check that these work and that $n=2,3,4,5,6,8,9,10$ don't work, so suppose FTSOC that $n\ge 11$ works. Let the three largest primes less than or equal to $n$ be $p<q<r$. Then since $r$ divides the product, it must divide $s+t$ for primes $s<t\le n$. But $s+t<2t\le 2r$, so $s+t=r$, so $s=2$, meaning that $q$ and $r$ are twin primes. But then $q$ divides $s+t$ for primes $s<t\le n$. Obviously this is impossible if $s=q$, and $q-2$ is not a prime(since then one of $p,q,r$ is a multiple of $3$), so $s+t<2q$, so $s+t=q$ too, contradiction.

The answer is $n=1,7$, easy to verify. We claim that if $n$ is a prime greater than $7$, this property will not hold. First, we consider the case when $n$ cannot be written in the form $2+p$ for any prime $p$. In this case, \[ \prod_{ p<q\le n, p,q, \, \text{primes}} (p+q) \]only contains terms less than $2n-1$; furthermore, it will not contain the term $n$, since any two not-equal-to-$2$ primes sum to an even number. So, of course, it is not a multiple of $n$, which is a contradiction. If $n$ can be written in the form $2+p$ for some prime $p$, we claim that the product is not a multiple of $p$. Because we are considering $n>7$, it follows that $p>5$. The maximum possible value of one of these terms in the product is $3+2p$ which is less than $3p$. So, we just have to prove that none of these terms are equal to $2p$ or $p$. If, FTSOC, a term is equal to $2p$, the two primes that sum to $2p$ must be $2+p$ and $p-2$, since the primes are greater than $n=2+p$. However, it is impossible for $p-2$, $p$ and $n=p+2$ to all be prime, since one of them will be a multiple of $3$; this is, unless one of these terms is equal to $3$ herself. But since $p > 5$, none of these terms are equal to $3$. We have a contradiction. If, FTSOC, a term is equal to $p$, due to parity, the two primes that sum to it are $2$ and $p-2$. But as established before, $p-2$ is not prime, so we have a contradiction. So, $n>7$ cannot be prime. But if we go from a prime $n$ to a composite $n$, only the left hand side increases, which will not cure indivisibilityness. So, no composites greater than $11$ work either. We check all remaining integers from $1$ to $10$, and confirm that these do not work either, hence $1$ and $7$ are the only solutions.

ISL 2022 N2. Find all positive integes $n$ such that \[n!\mid\prod_{p<q\le n,\,p,\,q \text{ are primes}}(p+q).\] Solution. The only solutions are given by $n\in\{1, 7\}$. Obviously, $n=1$ satisfies the problem condition. For $n=7$, we compute \[\prod_{p<q\le n,\,p,\,q \text{ are primes}}(p+q)=302400=60\cdot n!.\]We show that no other numbers satisfy the condition. Suppose $n\ge 2$. Choose the greatest prime $p\le n$. We have \[\prod_{p<q\le n,\,p,\,q \text{ are primes}}(p+q),\]so there exist two primes $p_1<p_2\le n$ such that \[p\mid p_1+p_2<2p\implies p_1+p_2=p\implies (p_1, p_2)=(2, p-2\ne 2).\]We also have \[p_2\mid n!\mid\prod_{p<q\le n,\,p,\,q,\,\text{primes}}(p+q),\]so there exist two primes $p_3<p_4\le n$ such that \[p_2\mid p_3+p_4<2p=2(p_2+2)<3p_2\implies p_3+p_4\in\{p_2, 2p_2\}.\] Case 1. $p_3+p_4=p_2$. Then $(p_3, p_4)=(2, p_2-2=p-4)$. Now, $p$, $p-2=p_2$, $p-4=p_4$ are all primes but $1$ of them is $0$ modulo $3$. The only possibility is $p-4=3$ and $p=7$. Case 2. $p_3+p_4=2p_2$. If $p_4<p=p_2+2$ then $p_4\le p_2$ and then $p_3+p_4<p_2+p_2=2p_2$, a contradiction. Thus $n\ge p_4\ge p$, but $p$ is the largest prime $\le n$, so $p_4=p=p_2+2$. Then $p_3=p_2-2$. Now, $p_2+2=p_4$, $p_2$, $p_2-2=p_3$ are all primes but $1$ of them is $0$ modulo $3$. This means the only possibility is $p_2-2=3$ and so $p=p_2+2=7$. The result follows.

Aryan-23 wrote: I claim that only $n=7$ works and it's easy to verify. Let $p$ be the largest prime atmost $n$. Since $p\mid n!$, there exists two primes $p_1,p_2$, such that $p\mid p_1+p_2<2p$. Hence $p_1+p_2=p$. Since $p$ is odd, $p_1=2$. Call $p_2=q$ from now on. Note similarly, $q\mid n!$, so there are primes $r,r'$ such that. $q\mid r+r'< 3q$. We consider two cases: If $r+r'=q$, then as before we get $r'=2$ and $r=q-2$. If $r+r'=2q$, then one of them, say $r'$ must equal $p$ for size reasons, so we again get $r=q-2$. In both cases we see that $r,r+2,r+4$ are all primes, and one of them must be divisible by $3$. This proves $r+4=p=7$. One can verify that $n=8,9,10$ don't work. Done. $\square$ I think you didn't do the case when $r=2q-p$.

David_Kim_0202 wrote: Aryan-23 wrote: I claim that only $n=7$ works and it's easy to verify. Let $p$ be the largest prime atmost $n$. Since $p\mid n!$, there exists two primes $p_1,p_2$, such that $p\mid p_1+p_2<2p$. Hence $p_1+p_2=p$. Since $p$ is odd, $p_1=2$. Call $p_2=q$ from now on. Note similarly, $q\mid n!$, so there are primes $r,r'$ such that. $q\mid r+r'< 3q$. We consider two cases: If $r+r'=q$, then as before we get $r'=2$ and $r=q-2$. If $r+r'=2q$, then one of them, say $r'$ must equal $p$ for size reasons, so we again get $r=q-2$. In both cases we see that $r,r+2,r+4$ are all primes, and one of them must be divisible by $3$. This proves $r+4=p=7$. One can verify that $n=8,9,10$ don't work. Done. $\square$ I think you didn't do the case when $r=2q-p$. Oh if you do smth at the middle of this you get this nvmd

Note that $1$ and $7$ are the only numbers to work which are less than $11$. If $n\ge 11$, let $x$ be the largest prime less than or equal to $n$. Since $x$ divides the product, there exists primes $p<q$ such that $p+q\mid x$. However, $p+q<2q\le 2x$ so $p+q=x$. However, since $x$ is odd, $p=2$ so $q=x-2$. Indeed, since $q\le n$, there exists primes $r<s\le n$ such that $q\mid r+s$. If $r+s=q$ then we get the same issue: $r+2$ and $s=q-2$ so $q-2,q,q+2$ are all primes, absurd. Thus, $r+s=2q$ so $s>q$ which implies $s=x$ and $r=q-2$, so $q-2,q,q+2$ are still all primes. We are done.

Aryan-23 wrote: Find all positive integes $n$ such that $$ n! \mid \prod_{ p<q\le n, p,q, \, \text{primes}} (p+q)$$ i think in the SL booklet it states $n>2$

Beautiful problem in my opinion. The answer ist $n=7$ only, which works. Note that if $n=r>7$ is a prime and $r-2$ is not then there does not exist a pair $(p,q)$ such that $p+q=kr$ as $p+q$ is even and less than $2r$, hence $r\mid r!$ but $r$ does not divide the RHS, a contradiction. Also letting $r'$ the smallest prime larger then $r$, we see that all the numbers $r,r+1,\ldots,r'-1$ also don't satisfy the problem condition as the RHS does not gain any new prime factors while $r$ still divides the LHS. Now, if $r'-2$ is not prime, we repeat the same process again. Otherwise $r'-2$ is prime (i.e. $r'-2=r$). Then, taking a close look the the RHS, we gain the prime factors \[(2+r')(3+r')\ldots(r+r')=(4+r)(5+r)\ldots(r*+r')(2r+2)\]where $r*$ is the largest prime less than $r$, which by assumption implies $r*\le r-3$, so the second largest factor is $r*+r'\le r-3+r+2=r-1$, which is also not divisble by $r$. Clearly, $2r+2=2(r+1)$ is also not divisible by $r$ for $r>2$. Thus for $n=r'$ the LHS is divisble by $r$ by the RHS is not, a contradiction again. Now let $r''$ be the smallest prime bigger than $r'$. As we do not gain any new prime factors on the RHS, while the LHS is still divisble by $r$, all the number $r,r+1,\ldots, r''-1$ do all not satisfy the problem condition. Now note that we cannot have $r''-2=r'$, as then one of $r,r',r''$, i.e. one of $r,r+2,r+4$ is divisible by $3$, impossible. Hence $r''$ is a prime such that $r''-2$ is not, so we may apply the same argument again. Doing this repeatedly, this shows that there are no solutions for $n\ge r$. Notice that $11$ is prime but $11-2$ is not, so there are no solutions for $n\ge 11$. Checking the remaining cases shows that $n=7$ is the only solution.

We claim, that $n=\boxed{7}$, is the only solution. This is easy to find out. Now, we check that $2,3,4,5,6,7,8,10$, don't work. So FTSOC we assume that there exists $n>10$, that works. Let the three primes, be $p_1, p_2, p_3$, so now, WLOG, assume that, $p_1<p_2<p_3.$ oops i cant solve the rest

If $n$ and $m$ are consecutive primes and a number between $n$ and $m-1$ dos not work, then anything larger than the number also does not work because the factorial is multiplied but the prime products stay the same. Testing $3,5,7$ results in $7$ working. But $8$ does not work, so anything between $8$ and $10$ also does not work. Let $n$ be a prime number that is more than $ 7$ Let $m$ be the first odd number less than $n$ that is not prime. $m \ge n-4$ because $n, n-2$, and $n-4$ are different mod $3$, so one of them has to divide $3$ by pigeonhole. The one that divides $3$ cannot be prime because $n \ge 11$, so $9$ is an odd that is divisible by $3$ that is not prime, so it would appear before $3$. To divide $m+2$, the two primes have to sum to $2m+4$ because two primes only sum to an odd if one is $2$, so the other one has to be $m$, but $m$ is not prime, so this would not work. If the sum is larger, then it would become too big because $3m+6>2(m+4)=2n$ for all positive $m$. If $m=n-4$, then the largest sum of two primes is $n+(n-2)=(m+4)+(m+2)=2m+6$. The second largest sum has a maximum value of $n+(n-6)=(m+4)+(m-2)=2m+2$ when $n-6$ is prime, which is less than $2m+4$, so it is impossible to reach $2m+4$. If $m=n-2$, then the largest sum of two primes is $n+(n-4)=(m+2)+(m-2)=2m$, which mappens when $n-4$ is prime. This is not large enough, so it is impossible to sum to $2m+4$ So when $n$ is more than $7$, there are no solutions, and the only solution is $\boxed{7}$.

Claim: $n \geq 11$ do not work Proof: Let the largest prime $\leq n$ be $r$ If $r-2$ is not a prime, consider the value of $V_r( \prod_{p<q \leq n} (p+q))$ Firstly, note that $p+q<2r$ which means that for $r|(p+q)$ , we must have $p+q=r$ Since $r$ is odd it implies that $p=2$ and therefore $q=r-2$ Which gives us a contradiction If $r-2$ is a prime, it implies that $r-4$ is not a prime since $3|r-4$ This time, we instead consider the value of $V_{r-2}( \prod_{p<q \leq n} (p+q))$ Firstly, note that $p+q<r+r-4=2(r-2)$ Similarly, this implies that $r-2|p+q$ which in turn means that $r-2=p+q$ Since $r-2$ is odd, $p=2$ and $q=r-4$ which gives us a contradiction Hence our claim must be true. Now we finish by checking all $n \leq 10$ which is fairly standard and you will find out that $n=7$ is the only solution I'm not sure whether to include $n=1$ but whatever

Shortlist 2022 N2 wrote: Find all positive integers $n>2$ such that $$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$ The only solutions are $\boxed{1,7}$ which is easy to check. Note that by $\pmod 3$ there doesn't exist $k \not =7$ such that $k,k-2,k-4$ are all prime.Now consider the largest prime $p$ less than $n$ for $n\geq 11$. If $p-2$ is not a prime then for any $2$ primes $q,r$ less than $n$ not equal to $p$,$p\neq q+r<2p$ since if $q,r$ are both odd then $q+r$ is even and since $p-2$ isn't a prime $2+p-2=p$ cannot be considered in the product and for any other prime $q,2+q \neq p$. If $p-2$ is a prime $p-4$ cannot be a prime,so $2+p-4=p-2$ or $p+p-4=2(p-2)$ cannot be considered in our product.For any other odd prime $q,q+2 \neq p-2,q+p \neq 2(p-2)$ or any other pair of odd prime $(q,r)$ both less than $p-2, 2p-4>q+r\neq p-2.p-2 \neq q+r$ since $q+r$ is even while $p-2$ is odd. Thus in the $1$st case $p$ divides no term in the product and in $2$nd case $p-2$ divides no term in product thus combining both cases we have $n! \not | \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$ for $n \geq 11$ .For $n \leq 10$ the cases could be checked manually to get that $n=1,7$ $\blacksquare$

yay I solved some real NT (prolly really basic for others but ok -_- ) Let $q,p,r$ be the largest primes such that $r < p < q \leq n$. If the condition is true then $$q! \ | \prod_{ p<q\le n, p,q \, \text{primes}} (p+q) = \prod_{ 2<q\le n, q \text{ primes}} (2+q) \prod_{ 3 \leq p<q\le n, p,q \, \text{primes}} (p+q)$$Notice that the all the terms in the second part of the product are even. Now since $q>2$, we can factor out all $2$'s in the product, which gives us $p<\frac{p+q}{2}<q$, which means that the largest prime factor in the second part of the product is less than $q$, which means $$q | \prod_{ 2<q\le n, q \text{ primes}} (2+q)$$which directly gives us $q=p+2$. Now using the same logic for $p$ we get that either $p=r+2$ or $p=\frac{q+r}{2}$, both of which correspond to the same result. Therefore the largest primes less than $n$ are of the form $p,p+2,p+4$.This is only possible for the triplets $(3,5,7)$, since one of $p,p+2,p+4$ is always divisible by $3$. Therefore the only possible values of $n$ are $7,8,9,10$ and it can be checked that only $n=7$ works $\blacksquare$

Let $$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)=S$$Claim- No prime $n$ other than $7$ satisfy. Proof- For the condition to satisfy there must be one term in $S$ which is equal to $n$ as $2n$ is not achievable. As $p_i,q_i,n$ all can't be odd therefore one of $p_i,q_i$ is $2$. Let $p_i=2$, thus $q_i= n-2$. Therefore $n,n-2$ both should be prime for this. This implies there are also prime $p_j,q_j$ such that $n-2 | p_j+q_j$ but $3n-6<n$ implies $n<3$ so we won't consider the condition. Therefore either $p_j+q_j = n-2 $ or $2n-4$. If it's $n-2$ it implies $n-4$ must also be prime. For the case when $p_j+q_j =2n-4$ consider both $p_j, q_j <n$ then $p_j+q_j$ can only be $2n-1,2n-2,2n-3,2n-4$. Therefore as both are distinct $(n-1, n-2)$ is the only possible pair but that also contradicts as both can't be prime. Hence, one of $p_j,q_j$ has to be $n$ in this case which again leads to the condition that $n-4$ is prime. We conclude $n,n-2,n-4$ are primes which is only possible for $n=7$. Claim- No composite $n$ satisfies. Proof- Say $t$ is an odd prime and $t+k$ be the next prime after $t$. Then $S$ is same for all $n$ equals $t,t+1,t+2 \cdots ,t+k-1$. Hence if $t$ doesn't satisfy neither will $t,t+1...$ satisfy. Hence the only cases to check should be $7<n<11$ of which no $n$ satisfies. Thus, only $n=7$ works.

Not a very difficult problem. I found it to be quite cute! Man... I really missed the final part of 3 primes in arithmetic progression with common difference $2$ means one of them have to $3$... Besides that, it was fun! Solution: The answer is $n = 7$ which works trivially. We will now head for the other direction. Denote $S = \{p_1, p_2, \dotsc, p_{k-1}, p_k(\coloneqq P)\}$ to be the set of all primes less than $n$ such that for $i < j$, we have $p_i < p_j$. Claim: $P - 2 = p_{k-1}$ and $p_{k-1} - 2 = p_{k-2}$. Proof: Since $P \le n$, $P$ must divide $n!$. This implies \[P \,\,\Bigg|\, \prod_{p < q\le n, p,q \text{ primes}} (p+q).\]Since $P$ is a prime, it must divide one of the paranthesis individually, say $(p_i + p_j)$ for $i > j$. It is clear that $p_i + p_j < 2P$, so we must have $p_i + p_j = P$. By parity reasons, we must have $p_j = 2$. Observe if $i < k-1$, then $p_i + 2 = P$ is impossible(intuitively, this is like skipping a prime). Therefore we have proven the first part. For the second part, we adopt a similar strategy. Rather than looking at $P$, we look at $p_{k-1}$. We again have \[p_{k-1} \,\,\Bigg|\, \prod_{p < q\le n, p,q \text{ primes}} (p+q).\]It is easy to see with $P-2 = p_{k-1}$ that none of the paranthesis can be greater than $2p_{k-1}$. Very similarly we can again conclude that $p_{k-1} - 2 = p_{k-2}$. This proves the entire claim. $\square$ We can finish easily from here. The claim gives us that there is a prime $q(= p_{k-2})$ such that $q, q+2, q+4$ is a prime. Looking at these numbers modulo $3$, we find that one of them will be always divisible by $3$. From here, its obvious that the only possibility for all of them to be prime is when $q = 3 \implies P = 7$. Since $P = 7$, we clearly get that $n < 11$. Upon checking the handful cases manually, we find that the claimed solution is the only possible one. $\blacksquare$

We claim the answer is $n=7$. Order the primes as $p_1,p_2,\dots$. Claim: The only thing that works for $n=\{3,\dots,10\}$ is $n=7$ Proof: We can easily see this by plugging in the values $\square$ Claim: There are no primes that work for $n\geq 11$. Proof: FTSOC, assume this works. Let $n=p$. If the prime right before $p$ is $p-4$, then the sequence is: \[2,3,\dots,p-4,p.\]However, this means the product will never have a factor of $p$ in it, meaning $p!$ will not divide the product. Therefore, the prime right before $p$ is $p-2$. If the prime before that is $p-6$, then the sequence is: \[2,3,\dots,p-6,p-2,p.\]However, this means that the product will never have a factor of $p-2$ because $2p-6<2(p-2)$, and $p-2$ itself cannot be formed as the sum of any two primes. Therefore, the prime right before $p-2$ is $p-4$. However, this implies that at least one of $p-4,p-2,p$ is $0\pmod{3}$, and since $p\geq 11$, this is a contradiction $\square$ Claim: If $n=p_1$ does not work, then $n=\{p_1+1,\dots,p_2-1\}$ does not work. Proof: Clearly, for all these composite integers: \[\prod_{ p<q\leq n} (p+q)\]is constant. However, $n!$ increases after $n=p_1$. Therefore, all these composite integers will not work if $n=p_1$ does not work $\blacksquare$

Solved with a_smart_alecks. We claim that the only solution is $n = 7$. We can manually check that $n < 11$ does not work. To show that all other values do not work, let $P$ be the greatest prime less than or equal to $n$, $Q$ the second greatest, and $R$ the third greatest. Observe that $P$ divides the LHS, and thus must divide the RHS too. In other words, there must exist two distinct primes less than or equal to $n$ that sum to a multiple of $P$. Since $P$ is the greatest prime less than $n$, we conclude that no two primes can sum to $2P$, so they must sum to $P$. Therefore, one of the primes must be even, and $P-2 = Q$. Additionally, $Q$ divides the LHS, and thus must divide the RHS too. Therefore, two distinct primes less than or equal to $n$ must sum to either $Q$ or $2Q$. In the former case, $Q-2 = R$ because one of the primes must be even, and in the latter case, $P$ must be one of the primes in the sum, so $R = Q-2$ again. In other words, $P$, $Q$, and $R$ are in an arithmetic sequence with common difference $2$, so one must be a multiple of $3$. However, this implies that $R = 3$, which means that $Q = 5$, $P = 7$, and so $n < 11$ and we are done.

This one was kinda funny since i got the thing about the largest prime $P$ but not the second largest prime $Q$. so when math boy started going down the same path as me, i was like “this probably does nothing” and it did something.

I claim $\boxed{n = 7}$ is the only such integer. Let $p$ be the largest prime. Then we have that if $p_1 < p_2$ are primes, that \[p \mid p_1 + p_2 < 2p \implies p = p_1 + p_2 \implies p_1 = 2 \implies p - 2 \text{ is prime},\]as $p$ is odd, so $p_1$ must be even. Then we have that $p_2$ is the second largest prime, so we have that \[p_2 \mid q_1 + q_2 < 2p_2 \implies p_2 = q_1 + q_2 \implies q_1 = 2 \implies p - 4 \text{ is prime}.\]If, however, one of $q_1, q_2 = p$ (say $q_2 = p$), we derive that $p_2 = q_1 + 2$, and similarly obtain that $p - 4$ is prime. Hence we have that $(p, p_2, q_2) = (p, p - 2, p - 4)$ are primes. Now we must have $p - 4$ is divisible by $3$, so that $p = 7$. Now checking all values for $7\leq n < 11$ we have that indeed the only viable solution is when $\boxed{n = 7,}$ as was initially claimed. $\blacksquare$

We claim only $n=7$ works. Let $r$ be the largest prime that is $\leq n$. Suppose $r-2$ is not prime and $n \geq 5$ (so that there is at least $3$ primes smaller than or equal to $n$). Since $0 < p+q < 2r$, we have $p+q = r$, which implies (WLOG) that $p=2$ and $q = r-2$. But $r-2$ is not prime, a contradiction. We then manually check that $n < 5$ has no solutions. Now suppose $r-2$ is an odd prime. Thus, $r-2 \mid p+q$ for some $p,q$. Suppose that $p,q \leq r-2$, then $p+q < 2(r-2)$ (since $p\neq q$), and thus $p+q = r-2$. This implies (WLOG) that $p=2$, and $q = r-4$, forcing $r-4$ to be a prime. Otherwise, we can assume WLOG that $p=r$, then $p + q > r > r-2$, and $p+q < r + (r-2) \leq 3r-6$ when $r \geq 4$, therefore $p + q = 2(r-2)$. This means $q = r-4$, again forcing $r-4$ to be a prime. We are left with the case $r < 3$, but clearly, in this case, $r-2$ cannot be an odd prime, a contradiction. We thus conclude that $r-4$ is a prime as well. We now have $r-4,r-2,r$ primes, which is impossible by taking mod $3$,unless $r-4=3$, in which case $r=7$. This means $7 \leq n \leq 10$, and we manually check (using $v_p$) that only $n=7$ works.

Aryan-23 wrote: I claim that only $n=7$ works and it's easy to verify. Let $p$ be the largest prime atmost $n$. Since $p\mid n!$, there exists two primes $p_1,p_2$, such that $p\mid p_1+p_2<2p$. Hence $p_1+p_2=p$. Since $p$ is odd, $p_1=2$. Call $p_2=q$ from now on. Note similarly, $q\mid n!$, so there are primes $r,r'$ such that. $q\mid r+r'< 3q$. We consider two cases: If $r+r'=q$, then as before we get $r'=2$ and $r=q-2$. If $r+r'=2q$, then one of them, say $r'$ must equal $p$ for size reasons, so we again get $r=q-2$. In both cases we see that $r,r+2,r+4$ are all primes, and one of them must be divisible by $3$. This proves $r+4=p=7$. One can verify that $n=8,9,10$ don't work. Done. $\square$ Can I ask why there exists two primes $p_1,p_2$ such that $p \vert p_1+p_2 < 2p$ and exsits $r,r'$ are prime such that $q\vert r + r' < 3q$

truongphatt2668 wrote: Can I ask why there exists two primes $p_1,p_2$ such that $p \vert p_1+p_2 < 2p$ and exsits $r,r'$ are prime such that $q\vert r + r' < 3q$ Since $p$ is the largest prime $\le n$, we have $p\vert n!$, so $p$ divides the product on the right side of the problem condition. Thus, there are $2$ $p_1$ and $p_2$ such that $p\vert (p_1+p_2)$. WLOG assume $p_1<p_2$. Since we defined $p$ as the largest prime $\le n$, we must have $p_1<p_2\le p$ so $p_1+p_2 < 2p$. The next part is similar: we have that $q$ divides the product on the right side of our condition, so there exist primes $r$ and $r'$ such that $q\vert (r+r')$, and as before we have $r+r' < 2p$. Assume $q\ge 5$. Since $q = p-2$, we see that $r+r' < 2p = 2q+4 <3q$. This only fails when $q \ge 5$ is false, which corresponds to very few cases of small $n$ to check manually.

Let $p_0$ be the largest prime lesser than or equal to $n$. $\implies \exists$ $p$ & $q$ such that $$p_0|p+q$$However $$p<q<p_0 \implies p+q<2p_0 \implies p+q=p_0$$$$ p_0 \ge 3 \implies p=2$$$\implies2+q=p_0$ where $q$ is the second largest prime divisor of $n!$. Now there are primes $p_1$ & $p_2$ such that $p_1<p_2$ and $$q|p_1+p_2$$However $$p_1+p_2 \le q+(q+2)=2q+2$$$$ \implies p_1+p_2 \in \{q,2q\}$$If $p_1+p_2=q$ then $p_1$ must be $2$ and $p_2$ must be $q-2$. If $p_1+p_2=2q$ then $p_2$ must be $q+2$ and $p_1=q-2$ So, in both cases, $q-2$,$q$ and $q+2$ are all primes which implies $q-2=3$ due to mod $3$ so $p_0=7$ Hence, $n\in \{ 7,8,9,10 \}$ and by casework, we see $n=7$ is the only solution.

Only $\boxed{n=7}$ works: It is not hard to check the cases when $n<11$ so assume that $11\leq n$. Let $r$ be the largest prime that divides $n$. If $r-2$ is composite then there are no two primes at most $n$ (hence at most $r$) that sum to a multiple of $r$. If $r-2$ is prime then $r-4$ is not prime so there are no two primes at most $n$ that sum to a multiple of $r-2$.

Note that $n\geq 3$. Let $p$ be the largest prime $\leq n$. Since $p\mid n!$, it follows that there exists primes $p_1$ and $p_2$ such that $p\mid p_1+p_2$. Since $p_1,p_2<p$, we have $p=p_1+p_2$. Since $p$ is odd, it follows that $p_1=2$ and $p_2=p-2$. Thus $p-2$ is prime. Similarly, $p-4$ is also prime. Since one of $p,p-2,p-4$ is divisible by $3$, we must have $p=7$. Now we only have to check $n=7,8,9,10$. The only solution is $\boxed{n=7}$. $\square$

Observe that $f(n) = f(p_k)$ where $p_k$ is the closest prime to $n$ and number of primes $ [\leq n] = k$ Lets assume $ n \geq 11$ Observe that $11=2+9 \implies 11$ will have to be formed using $22$ on RHS Now $22 = 13+9 =15+7 = 17+5 \implies$ none of the numbers from $11$ to $16$ follow the property in question and $17$ might, but $17 = 2+15$ so to make RHS divisible by $17$ we will have to make $34$ $34 = 19+15 = 23+11 \implies n \geq 23$ but $23 = 2 + 21$ We prove that this will happen indefinitely. Observe that $11$ is a prime of the form $6K + 5$, so $11$ will be formed using $ 2 \cdot 11 = 2(6K+5)$ to write which we will need to find $2$ primes whose sum is $P(a)$ both of which are $ > 11$ $\implies$ the primes are $5+x, 5-x$ $ (mod 6)$ which can be primes only for $x = 0$, which means we ended up on another prime of the form $6k+5$ which is $>11$

Problem DIES Let $p$ be the biggest prime less than $n$ $\exists$ distinct primes $p_i$ and $p_j$ such that $p\mid p_i+p_j<2p \implies p_i+p_j=p$ since $p$ is odd assume $p_i$ to be even, so $p=2+p_j$, this gives us that $p-2$ is also a prime, note that $p_j$ will be the prime that precedes $p$. Repeating this process again for $p_j$ we have, $p_j\mid p_k+p_l\implies p_k+p_l < 2p_j\implies p_j=p_k+p_l=2+p_l\implies p-4$ i a prime. So we need $p,p-2,p-4$ all to be primes, taking $\pmod 3$ we get $\boxed{p=7}$ as the only solution. $\blacksquare$

for $n\ge 11$ prime, if $n-2$ is not prime, then its not possible since $p+q=n$ with $p, q$ prime doesn't exist. when $n-2$ prime, then $n-4$ is not a prime(since $n \ge 11$). Hence there exist 2 primes $p,q \le n$ with $p+q=2n-4$, but that is a contradiction.

Nice! Let $n_{p}$ be the greatest prime $\leq n$. Then, note that $n_{p} \mid n!$. Since $$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q),$$we know that there exist primes $p,q$ such that $ n_{p} \mid p+q$. Since $p+q< 2n<2n_{p}$ (by Bertrand’s Postulate), $p+q=n_{p}$. Checking mod $2$ gives us that one of $p$ and $q$ is $2$, to, in other words, $n_{p}-2$ is also a prime. But then there also exist primes $r,s$ such that $r+s \mid n_{p}-2$. Since $r,s \leq n_{p}, n_{p}-2,$ $r+s\leq 2n_{p}-4$ with equality possible only when $n_{p}=7$. Checking $n=7,8,9,10$ we get $n=7$ as the only solution. $\blacksquare$

slight variation

See that if p be the maximum prime less than n≥5 and then p-2 has to be prime also so that the product is divisible by p now thus we get p-4 has to be prime also so by the claim that p=±1(mod 3) we get p=7 now we get n=3,4 or 7,8,9,10 now by checking these we get n=7 is the only solution is it not oral?

@same as above