PUMaC 2023 · 数论（B 组） · 第 6 题

6. What is the smallest possible sum of six distinct positive integers for which the sum of any five of them is prime? Proposed by Austen Mazenko Answer: 74 The smallest possible sum is 74, achieved for the integers 1 , 3 , 7 , 15 , 21 , 27. Consider the sum of the smallest five integers, which is 47 in this case. Suppose there was a more optimal solution with a smallest sum larger than 47. Now, the five other sums must be 2 distinct prime numbers greater than this value, meaning the sum of the largest five integers is at least six primes greater than 47; in particular, it’s at least 73, meaning the sum of all six integers is at least 74 as claimed. Now, suppose the sum of the smallest five integers is p < 47, and let the integers be a , a , ..., a 1 2 6 in increasing order. Thus, a + a + a + a + a = p , and since subbing out any of the summands 1 2 3 4 5 for a gives another prime sum, p + a − a is prime for 1 ≤ i ≤ 5. Since p ̸ = 2, it’s odd, 6 6 i so a − a is even meaning all the a have the same parity; hence, they’re all odd. Thus, 6 i i p ≥ 1 + 3 + 5 + 7 + 9 = 25, so we need to check p = 29 , 31 , 37 , 41 , 43. Define p := p + a − a , i 6 i so p is some prime larger than p . Summing the equation a − a = p − p for 1 ≤ i ≤ 5 gives i 6 i i P 5 P 5 p + ( p − p ) i i =1 5 a − p = ( p − p ), so a = . Since the a are increasing, p is the largest 6 i 6 i 1 i =1 5 difference, and thus a = a + p − p > p − p . For a fixed p , we maximize a by taking 6 1 1 1 1 6 P 5 p + ( p − p ) i i =1 p , p , p , p to be the four primes just less than p . Then, we need > p − p , 2 3 4 5 1 1 5 which is equivalent to P 5 5 X p + ( p − p + p − p ) i 1 1 i =1

p − p ⇔ p > ( p − p ) ⋆ . 1 1 i 5 i =1 Looking at differences of consecutive primes at least 29 and at most 73, namely 2 , 6 , 4 , 2 , 4 , 6 , 6 , 2 , 6 , 4 , 2, P 5 we see that the minimal possible value for ( p − p ) is 2+(2+4)+(2+4+6)+(2+4+6+2) = 1 i i =1 P 5 34, so p = 29 , 31 do not work. Finally, since p + ( p − p ) must be a multiple of 5, sim- i i =1 ply checking p = 37 , 41 , 43 gives the result. Specifically, if p = 43, then p = 67 or 71. To 1 accommodate the modulo 5 constraint, the only values for the other p are 47 , 53 , 59 , 61 and i 47 , 53 , 59 , 67, respectively, but these do not satisfy ⋆ . If p = 41, then to accommodate the modulo 5 constraint either p = 71 in which case the other primes must be 43 , 47 , 61 , 67 or 1 47 , 53 , 61 , 67, neither of which satisfy ⋆ , or p = 67 so the other primes are 43 , 47 , 53 , 59 which 1 also fail. Finally, if p = 37, then in order for ⋆ to be satisfied, we must have p = 71 and the 1 other primes are 53 , 59 , 61 , 67, but this violates the modulo 5 condition, so we may conclude.