208 PHILOSOPHICAL TRANSACTIONS. [anNO 1667. 



resolved long since by John Geysius, by the help of particular multipliers, such 

 as those above mentioned, and published by Alsted in his Encyclopaedia, 

 An. ] 630, and by Van Schooten in his Miscellanies. 



We shall clear up what authors have omitted concerning the definition and 

 demonsti^tion of such fixed multipliers, &c. And therefore say, that each 

 multiplier is relative to the divisor to which it belongs, and thus define it : 



It is such a number as, divided by the rest of the divisors or their pro- 

 duct, the remainder is O, but, divided by its own divisor, the remainder is 

 an unit. 



We required the divisors proposed to be primitive to each other, /. e. that 

 no two or more of them can be reduced to less terms by any common divisor. 

 For if so, the question may be possible in itself, but not resolvable by help of 

 such multipliers, such being impossible to be found. The reason is, because 

 the product of an odd and an even number is always even, and that divided by 

 an even number, leaves either nothing or an even number. 



28. r4845 



Divisors 1 9 > The multipliers relative to them are < 4200 

 13J '^6916 



The difinition affords light enough for the discovery of these numbers. To 

 instance in the first ; the product of 19 and 15 is 285, which multiply by all 

 numbers successively, and divide by 28 till you find the remainder required. 

 Thus twice 285 is 570, which divided by 28, the remainder is 10 ; also thrice 

 285 is 855, which divided by 28, the remainder is 15. Thus if you try on 

 successively, you will find that 17 times 285, which is 4845, is the number 

 required, which divided by 28, the remainder is an unit. Hence then we shall 

 find that 



4845^1 rl9, 15, 17- 



4200 > is equal to the solid or product of < 28, 15, 10. 

 6916^ ^28, 19, 13. 



More easy ways of performing this postulatum are to be found in Van 

 Schooten's Miscellanies, and Tacquet's Arithmetic. 



For illustration of the rule proposed, take this example. 



Products. 



J , Solar cycle 25•^ 4845-j 121125 



,f{^^ F^^ Lunar cycle \6 >The multipliers 4200 > 67200 



1008, the jndiction 6^ 6916'' 41 496 



The sum of the products 22982 1 

 which divided by 7980, the remainder is 6381, for the year of the Julian 

 period ; from which subtracting 709, there remains 5672 for the age of the 

 world, according to archbishop Usher. 



