By W. W. Rouse Ball

*The arithmetic Teacher.*Index. References for extra research. comprises a hundred and fifty black-and-white line illustrations.

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Enquire what was the number of the card chosen. Then, if you deal, and as soon as you come to the original first card begin (silently) to count, reckoning this as one, the selected card will appear at the number mentioned. Of course, if all the cards are dealt before reaching this number, you must turn the cards over and go on counting continuously. Another similar trick is performed by handing the pack face upwards to some one, and asking him to select a card and state its number, reckoning from the top; suppose it to be the nth.

On CA take a point H so that CH = CD. Bisect BD in K. Join HK and let it cut CB (produced) in L. Join DL. Bisect DL at M , and through M draw M O perpendicular to DL. Bisect HL at N , and through N draw N O perpendicular to HL. Since DL and HL intersect, therefore M O and N O will also intersect; moreover, since BDC is a right angle, M O and N O both slope away from DC and therefore they will meet on the side of DL remote from A. Join OC, OD, OH, OL. 40 SOME GEOMETRICAL QUESTIONS. [CH. II The triangles OM D and OM L are equal, hence OD = OL.

This has led some writers to think that Fermat used none but elementary algebraic methods. This may be so, but the following remark, which I believe is not generally known, rather points to the opposite conclusion. He had proposed, as a problem to the English mathematicians, to show that there was only one integral solution of the equation x2 + 2 = y 3 : the solution evidently being x = 5, y = 3. On this he has a note* to the effect that there was no difficulty in finding a solution in rational fractions, but that he had discovered an entirely new method—sane pulcherrima et subtilissima—which enabled him to solve such questions in integers.