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10 August, 11:31

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

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  1. 10 August, 13:00
    0
    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
  2. 10 August, 14:08
    0
    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.
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