LESSONS IN ARITHMETIC. 





LESSONS IN ARITHMETICS V. 



^ION. 



M . i iln.r.ii/ h'.w many tiuiea ono number ia con- 



ion. 



..led is called tlm Dividend. 

 uhich \vi> il.it-itl'- is railed tin- liiritOT. 



nuiiilMT i'l' tinier which tho Dividend 

 contain^ tho Divuor U called tho Quotient (Latin 



:l "). 



: placed between two numbers means that 1!. 



. the second. Tims, 19 4- 5 means 19 divided 

 by 5. 



If tho Dividend does not contain the Divisor an exact number 

 of times, it will contain it number of tim. 



it) with a number left over, which will bo less than thn 

 The number left over in this case is called tho 



Tims, when we nay that 5 is contained in 10 3 times and 4 

 uu dividend, the divisor, ;: the quotient, and 1 the 

 remainder. 



This fact may bo exhibited in tho following form : 

 19 - 3 x 5 + 4 



2. It will readily bo perceived that division is, in reality, only 

 a short method of performing a series of subtractions, in tho 

 same way as multiplication is a convenient method of perform- 

 ing a series of additions. For instance, to find how many times 

 5 is contained in 19, subtract 5 (tho divisor) continually from 

 19 (tho dividend), until tho number is exhausted, or a number 

 less than 5 is left ; then, counting tho number of these subtrac- 

 tions, we shall get the quotient. Thus, 5 from 19 leaves 14, 5 

 from 14 leaves 9, 5 from 9 leaves 4. Since 5 has been sub- 

 tracted 3 times from 19, leaving 4 as a remainder, we see that 

 19 divided by 5 has 3 for its quotient, leaving 4 as a remainder. 



N.B. It is evident, from the nature of division, that the 

 product of the quotient and divisor, added to tho remainder, ia 

 equal to the dividend. 



3. Metltad of Division. Tho method we are about to explain 

 depends upon the truth of the following principle : 



If the dividend be split up into any number of parts, of 

 which the sum is equal to tho dividend, then, if we divide each 

 part separately by the divisor, tho sum of all tho quotients so 

 obtained will be tho quotient required. 



For instance, 18 is equal to the sum of 9 and 6 and 3. The 

 quotients of these, divided respectively by 3, arc 3, 2, and 1, 

 which, added together, make 6, the quotient of 18 divided by 3. 



Similarly, 36 is 28 + 8, and therefore 36 divided by 4 is the 

 sum of tho separate quotients of 28 and 8 by 4, wlu'ch are 7 and 

 2 respectively. Hence 7 + 2, or 9, is the required quotient. 



It must be observed that if, the quotient of a given dividend 

 and divisor being known, the dividend be increased by annexing 

 any number of ciphers to it, tho new quotient is obtained by 

 annexing the same number of ciphers to tho quotient. Thus, 

 28 divided by 4 has tho quotient 7 ; and 28000 divided by 7 

 is 4000, 



4. To divide 5356 by 4. 



5356 5 thousands + 3 hundreds + 5 tens + 6 units. 



Now 5 contains 4 once, with remainder 1 : therefore 5 thousands 

 contain 4 one thousand times, with remainder 1 thousand. 



Add this remaining 1 thousand to the 3 hundreds, thus making 13 

 hundreds. 



Now 13 contains 4 three times, with remainder 1 ; therefore 13 hun- 

 dreds contain 4 three hundred times, with remainder 1 hundred. 



Add this remaining 1 hundred to the 5 tens, thus making 15 tens. 



Now 15 contains 4 three times, with remainder 3 : therefore 15 tens 

 contain 4 thirty times, with remainder 3 tens, or 30. 



Add this remaining 30 to the 6 units, thus making 36 units. 



Now S6 units contains 4 nine times. 



Therefore 1 thousand, 3 hundreds, three tens, and 9 units are the 

 number of times the parts into which 5356 has been divided contain 

 tho divisor 4 respectively. Their sum, therefore, is the required 

 quotient : this is 



1 thousand -t- 3 hundreds + 3 tens + 9 units, i.e. 1339. 



5. The above is tho analysis of the following shorter process, 

 and will bo seen fully to explain it : 



"Write down tho dividend and diviaor as in tho margin; 

 then say 4 in 5 is contained 1 time, with 1 over. Write 

 the quotient 1 under the 5, and placing the remaining 

 1 before the next figure of the dividend 3, say, 4 in 13 





i> contained 3 time* and 1 over. Write toe quotient 9 



under the Moond figure in the dividend, and prefixing the re 



. fay, -1 in 15 in contained 3 tame* and 3 over. 



.. quotient 3 under the third figure in the dividend, and 



I.rciixintf the ruiiiiiiiiinjf 3 to the 6, My, 4 in 30 U contained 9 



time*, with no remainder, and write down the 9 under the la*t 



r unit's figure of the dividend. 



It will be neon that when, to get the first figure of the 

 imi.ticnt, we say 4 in 5 U contained once, with remainder 1, w 

 really indicate that 4 is contained in 5000 1000 time*, v. 



>. \vhi--h 1000 we carry on to add to the next three 

 :.!, which really indicates 300, and HO on; a* will 

 be seen by comparing the process with the analysis of the methof 

 in Articlo 4. 



C. To divide 7499 by 9. 



.;; J 



Here, since 7, the first figure of tho dividend, is less than tb 

 divisor, 9, wo take tico figures of tho dividend, and say, 9 in 74 

 is contained 8 times, witli a remainder 2, and put down the 8 

 under the second figure of tho dividend (reckoning from 1 

 hand). Then, proceeding as in the previous example, we say, 

 9 in 29 is contained 3 times and 2 over ; and again, 9 

 contained 3 times and 2 over. This last 2 is 2 unit", and it 

 therefore tho remainder left after dividing 7499 by 9. It >! 

 generally written after the quotient, as above. 



This method, which is only conveniently applicable wL 

 divisor is a small number (generally one figure), is called Short 

 Division. 



EXERCISE 8. 



(1.) Divide 658 by 2 ; 537 by 3 ; and 7891011 by 6. 



(2.) Divide 4389127 by 8 ; 407792 by 11 ; and 5349279 by 9. 



(3.) Divide 41239789 by 12 ; and 54937862 by 5. 



(4.) Divide each of the numbers contained in the square im 

 Ex. 4, page 23, successively by 2, 3, 4, 5, t5, 7, 8, 9, 10, 11, 12. 



(5.) Divide each of the numbers contained in the square im 

 Ex. 4, page 23, successively by 2, 3. 4. 5. t5, 7, 8, 9, 10, 1 



(6.) Divide each of tho numbers 1010421690, 7689768432134, 

 54932684736856, and 428571428571496, by all the number* 

 from 2 to 12 inclusive. 



7. To divide 9298 by 35. 



Arrange the figures as in the margin ; then say, the largest 

 number of times which 35 is contained in 92 is 2 t'\gsBU'**t 

 times. Write the 2 on the right, to form the first 

 figure of tho quotient, and subtract 2 X 35 i.e., __ 

 70 from 92, leaving 22. Annex to this remainder 229 

 the next figure (9) in the dividend, thus making 210 

 it 229. 



Then say, the greatest number of times that 35 is 

 contained in 229 is 6 (which must be found by trial). 

 Put down the 6 to form the next figure in the 

 quotient, and subtract 6 times 35 i.e., 210 from 

 229, leaving a remainder 19. To this annex the last figure (8) 

 of the dividend, making it 198. 



Then say, the greatest number of times which 35 is contained 

 in 198 is 5. Write down the 5 to form the next figure in the 

 quotient, and subtract 5 times 35 i.e., 175 from 198, leaving 

 23. 265 is the required quotient, and 23 is the remain ; 



Henco 9298 = 265 X 35 + 23. 



8. A careful examination of tho above process will show that 

 what wo havo really done is equivalent to saying : 35 i 

 tained in 92 hundreds tn-o hundred times, with a remai: 

 hundred ; then, subtracting 200 times 35 i.e., 7 thousand-' 

 from 9298, we have 2298 left. 



Next we say : 35 is contained in 229 t^ns nifty time*, with a 

 remainder of 19 tens ; then subtracting 60 times 35 i.- 

 from 2298, we have 198 left. 



Next we say : 35 is contained /re times in 198, with a re- 

 mainder 23. 



Hence we see that after taking away 35, first, 200 times from 

 the dividend, again, 60 times from what is left, and again, 

 from what is left, wo have 23 units over, a number which ia lesi 

 than 35. 



Henco we see that 35 is contained in 9298 



300 -f 60 + 5 .., 263 times with a remainder 23. 



