Question

If p q r is prime numbers such that pq+r=73, what is the least possible value of p+q+r

Answer 1

The answer to this question would be: p+q+r = 2 + 17 + 39= 58

In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
Since 73 is an odd number, it is clear that one of the p q r needs to be an even number. 

There is only one odd prime number which is 2. If you put 2 in the r the result would be:
pq+2= 73
pq= 71
There will be no solution for pq since 71 is prime number. That mean 2 must be either p or q. Let say that 2 is p, then the equation would be: 2q + r= 73

The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
r= 73-2q
r= 73- 2(17)
r= 73-34=39

p+q+r = 2 + 17 + 39= 58

Answer 2

Note that if p and q are both odd and r is any prime apart from 2, the sum will always be even, because the sum of two odds is always even. This means that p (or q) must be 2. So 2q+r=73 and q+r=s-2, so q=s-2-r.
Divide through by 2: q+r/2=73/2. Also s-2-r+r/2=73/2 and s=77/2+r/2. We need minimum s. Therefore r=3 and s=40, q=35, but is not prime. Move to next prime after 3, which is 5, so p=2, s=41, r=5 and q=34, not prime. But as we step through the primes, s is increasing by 1 and q is decreasing by 1. Let q=31, which is prime, then r=11, also prime. So s=44, p=2, q=31, r=11 seems to be the solution for minimum s.