Without loss of generality we can assume $a>b>c$ <1> Clearly $a+b+c$ is the largest <2> It's easy to see that since <1> we get $a+b+c>a+b-c>a+c-b>b+c-a$ And now we compare $b+c-a$ with $c$; we have $c-(b+c-a)=a+c-(b+c)=a-b>0$ Thus $c>b+c-a$ and therefore $b+c-a$ is the smallest <3> Since <2> and <3> we get $a+b+c-(b+c-a)=d \rightarrow 2a=d$ Thus $d$ is maximal when $a$ is maximal. If $b+c-a=800-a \rightarrow \max a =797 \rightarrow d=1594$ If $a+b=800$ We know that there are no even number in $a,b,c$ (it's easy to prove) So $b\geq 3 \rightarrow a\le 797 \rightarrow d\le 1594$ If $a+c=800$ similarly, $c\geq 3 \rightarrow a\le 797 \rightarrow d\le 1594$ So $\boxed{d=1594}$ is the maximum

From the number theory marathon in the pre-olympiad forum: KittyOK wrote: Problem 337 The answer is $1594$ which can be achieved by, for example, $(a,b,c)=(13,787,797)$. It is easy to check that $a,b,c$ are all odd, hence $\ge 3$. Suppose wlog that $a+b=800$. Since $a+b-c>0$, $c<a+b=800$. Thus, $c\le 797$. The largest is $a+b+c\le 800+797=1597$. The smallest is $\ge 3$. So, $d\le 1597-3=1594$.

Initially I made this problem harder by failing to realize that $a+b+c$ is also a term in the sequence.Without loss of generality let $a > b >c$.Then some trivial checking gives that we should have either $a+b+c>a+b-c>a>b>a+c-b>c>b+c-a$ or $a+b+c>a+b-c>a>a+c-b>b>c>b+c-a$.So we obviously have to maximize $(a+b+c)-(b+c-a)=2a$.Now first note that $a$ cannot be greater than $800$ because since $b+c$ is the minimal two-sum and we should have $b+c \le 800$ and if $a$ were too exceed $800$ then $b+c-a$ were to be negative.Also note that $799=17 \cdot 47$ is composite.Thus $a \le 797$. Now the tedious part of the problem is showing that $a=797$ is attainable,if you take $b=787$ and $c=13$.(I guess all the terms of the sequence are primes then,if i'am quite efficient in the division process ).So my answer is $1594$.

horizon wrote: Let $a,b,c$ be positive integers such that $a,b,c,a+b-c,a+c-b,b+c-a,a+b+c$ are $7$ distinct primes. The sum of two of $a,b,c$ is $800$. If $d$ be the difference of the largest prime and the least prime among those $7$ primes, find the maximum value of $d$. I had also seen it in some AMC/AIME.

Proposed by Leung Tat Wing (HKG).

In fact the only possible value of the difference $d$ of the largest and smallest of these seven primes is 1594. Proof: The given conditions implies there exists an odd prime $p<800$ s.t. $p \in \{a,b,c\}$ and $(800 - p,800+p) = (q_1,q_2)$ are a pair primes. Let $r$ be the remainder of the division $d:3$. • If $r=0$, then $p=3$, yielding $q_2 = 800 +3= 803 = 11 \cdot 73$. • Futhermore, if $r=1$, then $3 \mid 800 + p = q_2$, which implies $800+p=3$, i.e. $p=-797$. • Finally, if $r=2$, then $3 \mid 800 - p = q_1$, which means $800-p=3$, i.e. $p = 800 - 3 = 797$. Consequently $p=797$, which clearly is the maximal value $M$ of the triple $(a,b,c)$ since $M \leq 800 - \min\{a,b,c\} = 800 - 3 = 797$.