CH. VII] MAGIC SQUARES 161 



to a square of the third order divided into nine cells. If a five- 

 shilling piece is placed in the middle cell c x and a florin in the 

 cell below it, namely, in a„ it is required to place the fewest 

 possible current English coins in the remaining seven cells so 

 that in each cell there is at least one coin, so that the total 

 value of the coins in every cell is different, and so that the sum 

 of the values of the coins in each row, column, and diagonal is 

 fifteen shillings : it will be found that thirteen additional coins 

 will suffice. Such problems or puzzles involve little or no know- 

 ledge of mathematics, and need no further mention here. 



ADDENDUM. 



Note. Page 161. Magic Coin Squares. The solution of the problem 

 enunciated in the text is as follows. Taking the notation of figure xxiv 

 we must put a double-florin and a sixpence in cellar, two double-florins 

 in cell 63, a half-crown in cell c 3 , a florin and a shilling in cell 63, a crown 

 and a florin in cell a it a crown and a half-crown in cell c 2 , a crown and 

 a sixpence in cell 61 : see The Strand Magazine, London, December, 1896, 

 pp. 720, 721. 



B. E. 



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