On the side of the box, 14 by 22+4/5, we can arrange 13 rows containing alternately 7 and 6 balls, or 85 in all. Above this we can place another layer consisting of 12 rows of 7 and 6 alternately, or a total of 78. In the length of 24+9/10 inches 15 such layers may be packed, the alternate layers containing 85 and 78 balls. Thus 8 times 85 added to 7 times 78 gives us 1,226 for the full contents of the box. 371.--GOLD PACKING IN RUSSIA. The box should be 100 inches by 100 inches by 11 inches deep, internal dimensions. We can lay flat at the bottom a row of eight slabs, lengthways, end to end, which will just fill one side, and nine of these rows will dispose of seventy-two slabs (all on the bottom), with a space left over on the bottom measuring 100 inches by 1 inch by 1 inch. Now make eleven depths of such seventy-two slabs, and we have packed 792, and have a space 100 inches by 1 inch by 11 inches deep. In this we may exactly pack the remaining eight slabs on edge, end to end. 372.--THE BARRELS OF HONEY. The only way in which the barrels could be equally divided among the three brothers, so that each should receive his 31/2 barrels of honey and his 7 barrels, is as follows:-- Full. Half-full. Empty. A 3 1 3 B 2 3 2 C 2 3 2 There is one other way in which the division could be made, were it not for the objection that all the brothers made to taking more than four barrels of the same description. Except for this difficulty, they might have given B his quantity in exactly the same way as A above, and then have left C one full barrel, five half-full barrels, and one empty barrel. It will thus be seen that in any case two brothers would have to receive their allowance in the same way. 373.--CROSSING THE STREAM. First, the two sons cross, and one returns Then the man crosses and the other son returns. Then both sons cross and one returns. Then the lady crosses and the other son returns Then the two sons cross and one of them returns for the dog. Eleven crossings in all. It would appear that no general rule can be given for solving these river-crossing puzzles. A formula can be found for a particular case (say on No. 375 or 376) that would apply to any number of individuals under the restricted conditions; but it is not of much use, for some little added stipulation will entirely upset it. As in the case of the measuring puzzles, we generally have to rely on individual ingenuity. 374.--CROSSING THE RIVER AXE. Here is the solution:-- | {J 5) | G T8 3 5 | ( J } | G T8 3 5 | {G 3) | JT8 53 | ( G } | JT8 53 | {J T) | G 8 J 5 | (T 3} | G 8 J 5 | {G 8) | T 3 G 8 | (J 5} | T G 8 | {J T) | 53 JT8 | ( G } | 53 JT8 | {G 3) | 5 G T8 3 | ( J } | 5 G T8 3 | {J 5) |