135.--THE STONEMASON'S PROBLEM.

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.)

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