Unfinished lines

Let's think of a simpler version of this problem where the expression given is only $(a_1-a_2)$ , and we want to find the largest $n$ such that we can guarantee there is some $a_1$ and some $a_2$ within the $12$ given primes such that $n|(a_1-a_2)$ . Let's also assume none of the given primes are factors of $n$ , and we'll take care of this assumption later.

There are only $\phi(n)$ possible values for $a_1 \pmod n$ , because $a_1$ is prime and we assumed it isn't a factor of $n$ . Therefore, if $12>\phi(n)$ , we can apply the pigeonhole principle and say that, within the given numbers, there are at least two congruent modulo $n$ , and, thus, we can always find two of these numbers such that $n|(a_1-a_2)$ . To maximize $n$ , we need to find the largest $n$ such that $\phi(n)<12$ . You can easily check that it's $30$ , with $\phi(30)=8$ . This means, from our assumption, that, as they're factors none of $2$ , $3$ , and $5$ can be among the given primes.

To solve the full problem, let's keep the assumption about the primes and deal with the special cases later. We know that we can guarantee $(a_1-a_2)$ is divisible by $30$ . This leaves us with $10$ primes, but still $10>\phi(30)$ , so we can guarantee $(a_5-a_6)$ is divisible by $30$ as well. Now we have $8$ primes left, but it turns out the only thing that can be said is that the sum can be made divisible by $2$ . So the answer is $30 \cdot 30 \cdot 2= \boxed{1800}$ .

It is important to see that all that has been said still works even if only $11$ primes were given, we would still be able to apply the pigeonhole principle in the required places. This means we won't have to deal with the special cases where only one of the factors of $30$ are present, we can just ignore it and pretend there are only $11$ primes.

Now lets cover the special cases, splitting them up in two:

• $2$ and $3$ are present.

Even though only 10 primes satisfy our conditions ( $2$ and $3$ don't satisfy because they're factors of $30$ ), we can still guarantee that $30|(a_1-a_2)$ , now, we can set $a_3=2$ and $a_4=3$ , to get $5|(a_3+a_4)$ . To make it divisible by $1800$ form here we only need $(a_5-a_6)$ to be divisible by $12$ , and we still have $8$ primes to use. This can always be guaranteed, as $8>\phi(12)$ .

• $5$ is present.

Note that if all $2$ , $3$ , and $5$ are present, we can just ignore the $5$ and look at the previous case, as everything still works with $11$ primes. Looking at the cases when $2$ and $5$ are present, or $3$ and $5$ are present, let's just ignore the $2$ or the $3$ , and pretend there are only $11$ primes, and $5$ is one of them. Also, we still have enough primes to guarantee that $a_1-a_2$ is divisible by $30$

If any of the primes remaning is congruent to $1 \pmod 3$ , say, $p$ , we can have $a_3=p$ and $a_4=5$ and $6|(a_3+a_4)$ . Now, to get to $1800$ , $(a_4-a_5)$ has to be divisible by $10$ , but we have at least $7$ primes left, and $7>\phi(10)$ .

If none of the primes remaining are congruent to $1 \pmod 3$ , there are only $\frac{\phi(30)}{2}=4$ possible values for $a_5 \pmod {30}$ . Since we have at least $8$ primes left (I'm not counting $5$ , and I'm considering we started with $11$ primes) , we can guarantee that $30|(a_5-a_6)$ , and you can choose any primes for $a_3$ and $a_4$ , and $(a_3+a_4)$ will be even.

The only thing left to explain is that the only thing you can say about the sum $(a_3+a_4)$ is that it's even. I'm pretty sure this is true, but I haven't had time to explore it further, I'd highly apreciate if anyone wants to contribute with this solution.

This is @Alapan Das 's solution, I am not the one to think of this solution, just post the solution here:

"Primes can have 1,3,7,9, as there last digits except 2and 5.Now using PHP we can say there are minimum three primes with same last term.So,if we take 6 primes then out of them we can subtract those primes with same last term.So,the given term is divisible by 100.Now as there are 12 primes, 11are definitely odd.Then the given term can divisible by 8.Thus te term is divisible by lcm(100,8)=200.Now we know any prime except 2,3 can be expressed as 6K ±1.So,we can easily understand logically that the given term is divisible by 72 .So its divisible by LCM(72,200)=1800."