The puzzle amounts to this. Find the smallest square number that may be expressed as the sum of more than three consecutive cubes, the cube 1 being barred. As more than three heaps were to be supplied, this condition shuts out the otherwise smallest answer, 23³ + 24³ + 25³ = 204². But it admits the answer, 25³ + 26³ + 27³ + 28³ + 29³ = 315². The correct answer, however, requires more heaps, but a smaller aggregate number of blocks. Here it is: 14³ + 15³ + ... up to 25³ inclusive, or twelve heaps in all, which, added together, make 97,344 blocks of stone that may be laid out to form a square 312 × 312. I will just remark that one key to the solution lies in what are called triangular numbers. (See pp. 13, 25, and 166.)
135.--THE STONEMASON'S PROBLEM.
From Amusements in Mathematics by Henry Ernest Dudeney.
