Can You Pile the Primes?

As a warm-up, suppose you split the first $N_2$ prime numbers into two groups with equal sums. The smallest value that works is $N_2 = 3$: the primes $2,3,5$ split as $\{5\}$ and $\{2,3\}$, each summing to $5$.

Your puzzle asks for three groups instead. Splitting the first $N_3$ primes into three groups with equal sums, what is the smallest $N_3$ for which this is possible?

The Fiddler, Zach Wissner-Gross, March 27, 2026(original post)

Solution

Equal thirds require the total to be divisible by $3$, so first find when the running sum of the primes is a multiple of $3$. The sums of the first one through nine primes are never divisible by $3$; the first time is at the tenth prime, $$2+3+5+7+11+13+17+19+23+29 = 129 = 3\cdot 43.$$ Divisibility is only necessary, so check that a split into three $43$s actually exists, and it does: $\{29,11,3\}$, $\{23,13,7\}$, $\{19,17,5,2\}$. Since no smaller count even clears the divisibility test, the answer is $$N_3 = \boxed{10}.$$

The computation

Add primes one at a time and, after each, test whether the list so far splits into three equal-sum groups by a backtracking placement (each prime dropped into a bin without exceeding the target third). The first count that succeeds is $10$.

from sympy import primerange
def can_split(nums, k):           # split the full list into k equal-sum groups?
    s = sum(nums)
    if s % k: return False
    t, nums = s // k, sorted(nums, reverse=True)
    if nums[0] > t: return False
    pots = [0] * k
    def place(i):
        if i == len(nums): return True
        seen = set()
        for j in range(k):
            if pots[j] in seen or pots[j] + nums[i] > t: continue
            seen.add(pots[j]); pots[j] += nums[i]
            if place(i + 1): return True
            pots[j] -= nums[i]
            if pots[j] == 0: break
        return False
    return place(0)
g = primerange(2, 10**6); P = []
while True:
    P.append(next(g))
    if can_split(P, 3): print(len(P)); break       # 10

Extra Credit

Now split the first $N_6$ primes into six groups with equal sums. What is the smallest $N_6$?

Solution

The same reasoning applies, with the total now needing to be a multiple of $6$ and each group large enough to hold the biggest prime. The first count meeting both conditions, and admitting an actual six-way split, is $$N_6 = \boxed{57}.$$ The first $57$ primes sum to $6870 = 6\cdot 1145$, and they partition into six groups of $1145$. No smaller count works.

The computation

Exhibit a split for the first $57$ primes: peel off one group summing to the target $1145$ at a time by a subset-sum pass, then confirm all six groups hit the target and nothing is left over.

def peel(nums, target):           # one subset summing to target
    reach = {0: []}
    for x in nums:
        for s in sorted([k for k in reach if k + x <= target], reverse=True):
            reach.setdefault(s + x, reach[s] + [x])
        if target in reach: break
    return reach.get(target)
P = list(primerange(2, 300))[:57]
t, pool, groups = sum(P) // 6, list(P), []
for _ in range(6):
    g = peel(pool, t); groups.append(g)
    for x in g: pool.remove(x)
print(t, all(sum(g) == t for g in groups), not pool)   # 1145 True True