354 Arithmetical Recreations. 



3, and not by 9, the period may not be divisible by 9, but it 

 may be by 3, and in this case divide the whole of the three 

 periods by 9, and the quotient will be the factor sought. 



Lastly, if the number given be divisible neither by 9 nor 

 by 3, divide by 9 the whole of 9 periods, and the quotient 

 will be the factor sought. 



If, for example, N be the number given — subject to the 

 condition of its being uneven, and not terminated by 5 — and we 

 convert $ into a periodic fraction, and if we represent the 

 period by P, we shall have 



J^ _P 



N 999 .... from whence we 



obtain 999 =PK 



If the number N is not divisible by 9, the period P will be 

 so. In this case let us call Q the quotient of P x 9, and we 



shall have 111 = 2 N, and we shall find the factor 



which is sought. 



As for the case in which N is a multiple of 3 or of 9, and 

 in which P may consequently be divisible by 9, it is plain 

 that we have only to take, instead of P, the reunion of 

 three or nine periods, so that the whole may be divisible 

 by 9. For example : suppose the number given be 7, we 



find f = 0,142,857, 142,857 .' ; the factor sought will 



be -^ = 15,873; and, in fact, 15,873 multiplied by 7, 

 gives 111,111 ; and twice 15,873, or 31,746, multiplied by 7, 

 gives 222,222, etc. 



If the number given be 3, we have ^ = 0,3333 ; 



and the period is divisible by 3, but not by 9, we therefore 

 divide by the reunion of three periods — that is to say, 333, 

 and the quotient 37 will be the factor required. 



If 27 be the number given, the period will be 037, which 

 is neither divisible by 9 nor 3, and consequently we must take 

 the reunion of nine of these periods to divide by 9, which gives 

 the number 411,522, 633,744, 855,967, 0,781,893. 



We see by this last example that for a small number given 

 as multiplicand, we may require a very high multiplier ; but, 

 on the other hand, a high multiplicand may require a small 

 multiplier, as, if the great number just recited had been that 

 given as a multiplicand, we should have found 27 as the 

 multiplier. 



