What is the smallest $N$ such that the set of the first $N$ cubes can be partitioned into $K$ subsets whose sums are equal?
(See the original framing in terms of golden orbs at fivethirtyeight)
My approach is not only computational, but pretty brute-force to boot: for each number $K$ of subsets, we step through the sets of the first $N$ cubes, and search the tree of partial assignments of cubes to partitions until we find a complete assignment to $K$ same-sum subsets.
On my PC, this code finds the solution for $K$ up to $8$ in about $72$ minutes: $1, 12, 23, 24, 24, 35, 41, 47$.
I’ll be curious to see if there are analytic approaches to this. I’m sure there are at least more efficient computational ones, given that puzzle author Dean Ballard has determined values up to at least $N = 10$.
import time # Returns True if arr can be # partitioned into K subsets with equal sum # Modified from a method found online: # https://www.geeksforgeeks.org/partition-set-k-subsets-equal-sum/ def isEqualSumPartitionPossible(arr, K): if (K == 1): return True # If total number of partitions is more than N, or # array sum is not divisible by K, then a partition # into equal-sum subsets is not possible N = len(arr) if (N < K): return False sum = 0 for i in range(N): sum += arr[i] if (sum % K != 0): return False targetSum = sum // K # initialize first subset sum as # last element of array and mark that as taken subsetSums =  * K taken =  * N subsetSums = arr[N - 1] taken[N - 1] = True # We avoid explicit recursion to hopefully save a # little runtime, using a state stack instead. # [Actual result: no time savings.] stack = [[subsetSums,taken,0,N-1]] while not stack ==  : subsetSums,taken,curIdx,limitIdx = stack.pop() if subsetSums[curIdx] == targetSum: # current index (K - 2) means we've found # (K - 1) subsets of sum targetSum. Untaken # numbers must then form the Kth. if (curIdx == K - 2): return True # otherwise start next partition stack.append([list(subsetSums),list(taken),curIdx+1,N-1]) continue # still adding to a partition. we start from limitIdx and add # untaken elements into current partition for i in range(limitIdx, -1, -1): if (taken[i]): continue taken[i] = True subsetSums[curIdx] += arr[i] # if no greater than targetSum, add task to stack if (subsetSums[curIdx] <= targetSum): stack.append([list(subsetSums),list(taken),curIdx,i-1]) # unmark the element and remove from sum as we move # to previous element in arr taken[i] = False subsetSums[curIdx] -= arr[i] return False # Main loop startTime = time.time() K = 0 while True: K += 1 arr =  N = 0 while True: N += 1 arr.append(N**3) if (isEqualSumPartitionPossible(arr, K)): print(K,N, "in", (time.time()-startTime)) break