28 ARITHMETIC. 



4. Subtract the last found product from "that part of 

 the dividend, under wliich it stands, and to the right 

 hand of the remainder bring down the next figure of 



the 



EXAMPLE. 



Divisor 36)85609 Dividendw 



isl part of the dividend 850QO 



36 X 2000=: 72000 20CO the I St quotient,, 



1st remainder - 13000 

 add 6co 



•Zd part of the dividend 13600 



36 X 300= 10800 - - 500 the 2d quotients 



2d remainder - 2S00 

 add 00 



3d part of th€ dividend 2800 



36 X 70= 2520 - - 70- the 3d quotier% 



3d remainder - 280 

 add 9 



4th pai't of the dividend 289 



36x8= 288 - - 8 the 4th quotient^ 



Last remainder ~ 1 2^178 sum of the quotients, 



• J ', or the anfw t^r. 



Explanation. It is evident, that the dividend is resolved 

 into these parts, 85000^-600-1-004-9 ;• for the first part of the: 

 dividend is considered only as S^y but yet it is truly 85000 ; and 

 therefore its quotient, instead of 2, is 2000, and the remainder 

 13000 ; and so of the rest, as may be seen in the operation. 



When there is no remainder to a division, the quotient is the 

 absolute and perfect ansvyer to the question ; but u'bere there is 

 a remainder, it may be observed, that it goes so much toward 

 another time, as it approaclies to the divisor : thus, if the remain- 

 der be a fpvirth part of the divisor, it will go one fourth of a time 

 more ; if half the divisor, it will go half of a time more ; and so 



on. 



