Posted on Mar 27, 2026
From Dean Ballard comes a premier puzzle of primes:
Suppose you want to make two groups with equal sums using the first \(N_2\) prime numbers. What is the smallest value of \(N_2\) for which you can do this?
The answer is three! (Clearly, that wasn't actually the puzzle.)
The first three primes are 2, 3, and 5, and you can split them up into two sets: \(\{2,3\}\) and \(\{5\}\). Sure enough, \(2+3=5\).
Your puzzle involves making three groups with equal sums using the first \(N_3\) prime numbers. What is the smallest value of \(N_3\) for which you can do this?
The total sum of the first \(N_3\) primes must be divisible by 3.
The first nine primes sum to \(2+3+5+7+11+13+17+19+23=100\), which is not divisible by 3. The first ten primes sum to \(129\), which is divisible by 3, so \(N_3 \ge 10\).
For \(N_3=10\), the target sum per group is \(129/3=43\). One valid partition is:
Each set sums to 43, so the smallest possible value is \(N_3=10\).
Now you want to make six groups with equal sums using the first \(N_6\) prime numbers. What is the smallest value of \(N_6\) for which you can do this?
Again, the total sum must be divisible by 6. The first time this happens is at \(N_6=57\).
The sum of the first 57 primes is \(6870\), so the target per group is \(6870/6=1145\). Here is one explicit partition into six groups of sum 1145:
Each set sums to 1145, so the smallest possible value is \(N_6=57\).