Divisible by 6

Consider 3 cases, when both of them are two, when exactly one of them is 2, and when none is 2.

Case 1: $2×2+3×3×4=40$ which is not divisible by $6$

Case 2: $2q+3×(q+1)×(q+2)$ now, since latter term is divisible by $6$ , $2q$ must be divisible by $6$ , thus $q=3$ .

Case 3: Note that both are odd in this case, thus former term is odd while latter is even, thus it cannot be divisible by $6$ .

So only possible solution(s) are from case 2, (2,3) and (3,2)

Hence $2+3+3+2=10$

First assume $p,q > 3$ . In that case, then $p,q\equiv 1,5\pmod{6}$ because they are primes.

This tells us that $pq\equiv -1, 1 \pmod{6}$

However, note that if either of $p,q\equiv 5\pmod{6}$ , then $(p+1)(q+1)(p+q)\equiv 0\pmod{6}$ but in conjunction with $pq\equiv -1, 1\pmod{6}$ we know that $pq+(p+1)(q+1)(p+q)\not\equiv 0\pmod{6}$

Thus, at least one of $p,q$ is $\le 3$ . Let's take this by cases.

First case: WLOG let $p=2$ . Then $pq+(p+1)(q+1)(p+q)=2q+3(q+1)(q+2)$

Since $2\mid (q+1)(q+2)$ , then $6\mid 3(q+1)(q+2)$ so we need $6\mid 2q$ which forces $q=3$ . This gives $(p,q)=(2,3)$ as one solution.

Second case: WLOG let $p=3$ . Then $pq+(p+1)(q+1)(p+q)=3q+4(q+1)(q+3)$

If $q > 2$ , then $3q$ is odd. $4(q+1)(q+3)$ is obviously even, so $3q+4(q+1)(q+3)$ is odd overall. But this means that it is not dividible by $6$ . Thus, $q=2$ is a possible answer. Indeed, plugging in $q=2$ gives $6\mid 66$ , a true statement. This gives the second solution, $(p,q)=(3,2)$

Thus, our answer is $2+3+3+2=\boxed{10}$

---

Perhaps an easier solution is as follows:

Note that $pq+(p+1)(q+1)(p+q)=(pq+p+q)(p+q+1)$ . If one of the variables isn't equal to $2$ , then $pq+p+q$ is odd, $p+q+1$ is odd, so $(pq+p+q)(p+q+1)$ is odd, therefore not divisible by $6$ . Thus, one of the primes is $2$ . WLOG let $p=2$ . Then our expression turns into $(3q+2)(q+3)$ . if $q=2$ , then the expression is $40$ which is not divisible by $6$ . Then, we know that $q+3$ is even and $3q+2$ is odd. Thus, either $6\mid q+3$ or $2\mid q+3\text{ and }3\mid 3q+2$ . Then second case is absurd since $3\mid 3q+2$ is impossible, so we must have $6\mid q+3$ which forces $q=3$ .

Thus, our solutions are $(p,q)=(2,3)\text{ or }(3,2)$

Let us first assume that $p,q$ are both odd numbers then,we see that $pq+[p+1][q+1][p+q]$ would be odd.But $6$ can't divide an odd number,thus,one of $p,q$ is $2$ .First,let us take that $p=2$ .By putting $p=2$ the given expression becomes, $2q+3[q+1][2+q].......................[1]$ .Now,we see that both $p$ and $q$ can't be equal to $2$ .Thus, $q$ is an odd number,say $2x+1$ .Putting this value in $[1]$ we get, $2[2x+1]+3[2x+2][2x+3]\\ =2[2x+1]+6[x+1][2x+3]$ .This implies that $6$ should just divide $2[2x+1]$ which implies that $2x+1$ should be divisible by $3$ .But this can only happen when $2x+1=3$ as it is a prime.Thus, $p=2$ and $q=3$ .But,there is one more solution which is $q=2$ and $p=3$ .Thus,sum $=10$ .

We first factor the expression as $6$ divides $(pq+p+q)\cdot(p+q+1)=A\cdot B$ . Now it's clear that $A$ can only be even when $p=q=2$ , which is not a solution. So $B$ must be even. If $p=2$ and $q$ is odd, then $A\cdot B=(3q+2)\cdot(q+3)=2q=0$ mod $3$ , so $q=3$ . Thus, $(2,3)$ and $(3,2)$ are solutions. Finally, we cannot have both $p$ and $q$ odd, for then $B$ would be odd. Therefore, we have $(2+3)+(3+2)=\boxed{10}$ .

If either p or q or both p and q are odd then the answer will be odd and not divisible by 6. So at least one of them needs to be even. So one of them needs to be 2. We can write : $\frac { 2q }{ 6 } +\frac { 3(q+1)(2+q) }{ 6 }$ = $\frac { q }{ 3 } +\frac { (q+1)(2+q) }{ 2 }$ Now (q+1)(2+q) can be divisible by 2 for any value of q . But for $\frac { q }{ 3 }$ to be an integer q needs to be divisible by 3 . So q can only be 3 . As we can interchange the values of q and p we got two solutions of (p,q) . (2,3) and (3,2) .

Hence 2+3+3+2=10

This was a fairly easy question.

Any prime number divided by 6, will leave a remainder of 2, 3, 1 or -1 Now, we can straight away eliminate the prime numbers which leaves a remainder of -1

As (-1+1) would be equal to 0, making the second product divisible by 6; but the first product can't be divisible by 6 in this case as only -1*0 can give us a remainder of 0 and no prime number divided by 6 has a remainder of 0. Thus, we can straight eliminate all the prime numbers which leaves a remainder of -1 when divided by 6

So we can check the remaining cases:

$2*3 + (2+1)*(3+1)*(2+3)$ is divisible by 6

$2*1 + (2+1)*(1+1)*(2+1)$ isn't divisible by 6

$3*1 + (3+1)*(1+1)*(3+1)$ isn't divisible by 6

So, the only possible pair of values would be which leaves remainders 2 & 3 when divided by 6, which are 2 & 3

As, we are looking for ordered pairs, we have two possible pairs (2,3) and (3,2) Adding all of which we have the answer as 10

First, let $x=p+q$ . Substituting, we get: $pq+(p+1)(q+1)(p+q)=(x+1)(x+pq)$

Case 1: $q=2$ $(x+1)(x+pq)=(p+3)(2p+p+2)=(p+3)(3p+2)$ It is easy to see that $6 \nmid 3p+3$ . This implies that $6 \mid p+3$ . We get: $p+3=6k \Rightarrow p=6k-3=3(2k-1)$ The only way this can be true is if $p=3$

Case 2: $p=2$

With similar logic to case 1 we get $q=3$

Case 3: $p \ne 2 \ne q$

Since $p$ and $q$ are odd:

• $x+1=p+q+1$ is odd

• $pq$ is odd

• $pq+x$ is odd

Therefore $(x+1)(pq+x)$ is odd and therefore indivisible by 6.

So $(p,q) \in \{(2,3),(3,2)\}$ and the required sum is $2+3+3+2=\boxed{10}$

Let's suppose that p,q>2. Then p,q are odd, the (p+1)(q+1)(p+q) is even and pq is odd. So 6 | an odd number (impossible). Thus one of p or q should be 2. Let's suppose, without losing the generality - the expression is simmetric - that q=2. Then 6 | 2q + 3(p + 1)(p + 2) = 3p^2 + 11p + 6 => 6 | p(3p + 11). That means 3 | p(3p + 11) => 3 | p or 3 | 3p+11. But 3 | 3p + 11 is impossible, so 3 | p => p = 3. Thus, the pairs are (2, 3) and (3, 2) => the asked sum is 10.

Step 1: $p,q\not\equiv -1 mod(6)$ otherwise pq will leave a remainder of $\pm$ 1 whilst the rest will leave a remainder of 0. Step 2: $p,q\equiv1mod(6) \Rightarrow\ pq + (p+1)(q+1)(p+q)\equiv3\not\equiv0 mod(6)$ Step 3: Now we must test the other primes, namely 2 and 3: $p\not\equiv\ qmod(6) \ as \ 2p^3 + 5p^2 +2p\not\equiv 0mod(6)$ (because even + odd + even = odd and I substituted p = q) Step 4: $\large p = 2 \ and \ q =3$ ... $pq + (p+1)(q+1)(p+q) = 66\equiv0 mod(6)$ Finally the same is true if we swap the values of p and q. $\large \therefore\displaystyle\sum_{i=1}^{n} p_i + q_i = 2+3+3+2 = \boxed{10}$