We can say of any number up to 100 whether it is possible or not to express it as the sum of two cubes, except 66. Students should read the Introduction to Lucas's _Théorie des Nombres_, p. xxx. Some years ago I published a solution for the case of 6 = (17/21)^3 + (37/21)^3, of which Legendre gave at some length a "proof" of impossibility; but I have since found that Lucas anticipated me in a communication to Sylvester. [Illustration] 21.--_The Ploughman's Puzzle._ The illustration shows how the sixteen trees might have been planted so as to form as many as fifteen straight rows with four trees in every row. This is in excess of what was for a long time believed to be the maximum number of rows possible; and though with our present knowledge I cannot rigorously demonstrate that fifteen rows cannot be beaten, I have a strong "pious opinion" that it is the highest number of rows obtainable. 22.--_The Franklin's Puzzle._ The answer to this puzzle is shown in the illustration, where the numbers on the sixteen bottles all add up to 30 in the ten straight directions. The trick consists in the fact that, although the six bottles (3, 5, 6, 9, 10, and 15) in which the flowers have been placed are not removed, yet the sixteen need not occupy exactly the same position on the table as before. The square is, in fact, formed one step further to the left. [Illustration] 23.--_The Squire's Puzzle._ The portrait may be drawn in a single line because it contains only two points at which an odd number of lines meet, but it is absolutely necessary to begin at one of these points and end at the other. One point is near the outer extremity of the King's left eye; the other is below it on the left cheek. 24.--_The Friar's Puzzle._ The five hundred silver pennies might have been placed in the four bags, in accordance with the stated conditions, in exactly 894,348 different ways. If there had been a thousand coins there would be 7,049,112 ways. It is a difficult problem in the partition of numbers. I have a single formula for the solution of any number of coins in the case of four bags, but it was extremely hard to construct, and the best method is to find the twelve separate formulas for the different congruences to the modulus 12. 25.--_The Parson's Puzzle._ [Illustration]