Pi Han's Triple Primes

We can rewrite the expression as $-(p-1)(q-1)(r-1)-1+pqr=2013$ which is equivalent to $(p-1)(q-1)(r-1)=pqr-2014$ Since we have an even expression on the left hand side, we must have an even expression on the right hand side, which can happen if and only if $p=2$ . Plugging $p=2$ into the original expression yields $qr+q+r=2015$ Adding 1 to both sides and we get $(q+1)(r+1)=2016$ Since $p$ , $q$ , and $r$ are distinct, we know that neither $q$ nor $r$ is 2 and that $(q+1)$ and $(r+1)$ are both even and have a factor of 2, which tightens our search a little bit. All pairs $((q+1),(r+1))$ such that $(q+1)(r+1)=2016$ and $q<r$ are as follows: $(2,1008), (4,504), (6,336), (8,252), (12, 168), (14,144),$ $(16,126), (18,112), (24, 84), (28,72), (36,56), (42,48)$

Checking these for $p, q$ prime shows the only prime values of $q$ that satisfy the equation are $3, 7, 11, 23,$ and $41$ . Thus, the answer is $3+7+11+23+41=85$

If $p,q,r$ are all odd prime numbers, then the left hand side of the equation is even which is contradict with 2013 is odd. So $p=2$ .

Substituting $p=2$ implies $qr+q+r=2015$ .

Which can be written as $(q+1)(r+1)=2016=2^5 \cdot 3^2 \cdot 7$ .

Consider all possible pairs of $q+1$ and $r+1$ where $q<r$ and both of them are even more than 2.

$(q+1,r+1)=(4,504),(6,336),(8,252),(12,168),(14,144),(16,126),(18,112),(24,84),(28,72),(36,56),(42,48)$ .

$(q,r)=(3,503),(5,335),(7,251),(11,167),(13,143),(15,125),(17,111),(23,83),(27,71),(35,55),(41,47)$ .

Choosing only $q$ and $r$ are prime numbers yields $(q,r)=(3,503),(7,251),(11,167),(23,83),(41,47)$ .

Hence the sum of all possible values of $q$ is $3+7+11+23+41=85$ .

When considering the parity of prime numbers, we know that almost all prime numbers are odd, but one prime number is even... $2$ ...which also happens to be the smallest prime. So in order to figure out if $2$ is going to equal p , we can consider the equation:

$pq + pr + qr - p - q - r = 2013$

when p is even and q and r are odd. We end up with:

$even + even + odd - even - odd - odd = odd$

which is in fact true.

Therefore, $2$ can be a solution for p .

If we consider all other cases, p , q , and r are all odd. If we plug in the parities into the equation, we get:

$odd + odd + odd - odd - odd - odd = odd$

which is not true. The right-hand side would actually equate to an even number.

Therefore, solutions only exist when $p = 2$ .

So by plugging in $2$ for p , the equation is now:

$2q + 2r + qr - 2 - q - r = 2013$ or $q + qr+ r = 2015$ .

Starting with $3$ , start plugging in prime values for q . We can then solve the equation for r , and if r is prime, we have a solution. The solutions end up being the triples: $(2, 3, 503), (2, 7, 251), (2, 11, 167), (2, 23, 83)$ , and $(2, 41, 47)$ .

After $q = 41$ , all solutions for r are either non-integer values or r < q which are not permitted.

Therefore, the sum of all possible q is $3 + 7 + 11 + 23 + 41$ which equals $85$ .

We can factorize the equation into p(q-1) + q(r-1) + r(p-1) . Hence, we see that if p,q and r are all odd, q -1, p -1 and r -1 are all even and the solution will be even. Therefore we conclude that p,q or r is even, and is this case since 2 is the only even prime and is the smallest prime, p =2. Substituting that value into the equation, we see that 2 q +2 r + qr -2- q - r = q + r + qr -2=2013. This means that q + r + qr +1=( q +1)( r +1)=2016. Then we find all factors of 2016, q +1 and r +1 that fulfill the condition that q and r are prime, also bearing in mind that q < r . We see that the possible values of q are 3,7,11,23 and 41. 3+7+11+23+41=85, which is how we obtain the answer.

First we check the parity on both sides of the problem....If $p,q,r$ are all odd primes then clearly the LHS is even but the RHS is odd....So clearly the value of the smallest prime has to be 2. Putting $p = 2$ in the equation we get we get $2q+2r+qr-2-q-r = 2013$ $q+r+qr = 2015$ $q(r+1)+(r+1) = 2016$ $(q+1)(r+1) = 2016$ we know $2016 = 2^{5}*3^{2}*7$ .So we will obtain $18$ pairs of values for $q$ and $r$ . Now we have to check all the possible pairs and determine which of the values obtain primes $q$ and $r$ .This is a tedious job but putting $q = 2k_1+1$ and $r = 2k_2+1$ may help a little....

Finally we obtain the following values of q for which r is also prime.They are $3,7,11,23,41$ .

Their sum is $\boxed{85}$ !!

If all the primes are odd, then the left side is even. So we must have $p = 2$ . Plugging in and simplifying yields $qr + q + r = 2015$ , or $(q+1)(r+1) = 2016$ . Now $2016 = 2^5 \cdot 3^2 \cdot 7$ has $36$ factors in $18$ pairs. Write them all down, subtract $1$ from each, test for primality. We get the solutions $(q,r) = (3,503), (7,251), (11,167), (23,83), (41,47)$ , which leads to an answer of $3+7+11+23+41 = \fbox{85}$ .