110 



THE POPULAB EDUCATOK. 



LESSONS 



ARITHMETIC. VII. 



3 



ABRIDGED METHODS OF MULTIPLICATION AND DIVISION 



(continued). 

 6. To divide by 10, or any power of 10. 



If the dividend have more ciphers for its right-hand figures 

 than occur in the power of 10 by which it is to be divided, we 

 need only take away from it the number of ciphers in the divisoi 

 to obtain the quotient. Thus, 873000 divided by 100 and 1000 

 respectively, gives quotients 8730 and 873. But suppose that 

 the dividend has no ciphers for its right-hand figures. Take, 

 for instance, the case of 87346 divided by 100. Cut off the two 

 right-hand figures viz., 46 from the dividend ; then 873 will 

 be the quotient and 46 the remainder. This is evident by 

 exhibiting the process analytically, thus : 

 87316 = 87300 + 46 



= 873 x 100 + 46 



Therefore 873 is the quotient, and 46 the remainder. 

 The same rule applies to dividing by any power of 10. 



7. Next, suppose the divisor to be not a power of 10, but to 

 have ciphers for its extreme right-hand figures ; for instance, to 

 divide 2764 by 300. There being hvo ciphers in 300, cut off 

 the two right-hand figures viz., 64 from the dividend, and 

 divide the 27 by 3 ; this gives 9, which will be the quotient, 

 and the 64 will be the remainder. This is evident by exhibiting 

 the process analytically, thus : 



2764 = 2700 + 64 



= 9 x 300 + 64 

 Therefore 9 is the quotient, and 61 the remainder. 



8. In this last case there is no remainder after dividing 27 

 by 3. But suppose we have 2964 to divide by 300 : 



Proceeding as before, cutting off the 

 <34 and dividing 29 by 3, we get a quotient 

 9 and a remainder 2. But evidently this 

 remainder is in reality 2 hundreds, or 

 200 ; and therefore, since 64 is also left over, the whole 

 remainder will be 264. Hence, in this case, any remainder 

 which is left must be prefixed to the figures cut off, in order to 

 give the whole remainder. The process is exhibited analytically 

 as follows : 



2964 = 2900 + 64 



= 2700 + 200 + 64 

 = 9 x 300 + 264 

 Hence 9 is the quotient, and 261 the remainder. 



9. We subjoin one other example : 



To divide 25329483 by 723000. 



Cutting off three figures, viz., 483, from the dividend since 

 there are three ciphers in the divisor we divide 25329 by 72J!, 

 by the common process of Long Division. This gives a quotient 

 35, and a remainder 24. Hence the required quotient is 35, and 

 the whole remainder will be got by prefixing the 24 to the 

 figures 483 cut off from the dividend. Hence the whole re- 

 mainder is 24483. The process is analytically exhibited as 

 follows : 



25329483 = 25329000 + 483 



= 35 x 723000 + 24000 + 483 

 = 35 x 723000 + 24483 

 Hence the quotient is 35, and the remainder 24483. 



EXERCISE 13. 



1. In one pound there are ten florins ; how many pounds are 

 there in 200 florins ? In 340 florins ? In 560 florins ? 



2. In one metre there are 100 centimetres; how many 

 metres are there in 65000 centimetres ? In 765000 centimetres ? 

 In 4320000 centimetres P 



3. Work the following sums in division : 



9264 remainder. 



3. 582367180309 -f- 100000000. 



4. 3360000 *- 17000. 



1. 26750000 -i- 100000. 



2. 144360791 * 1000000. 



4. How many vehicles at 70 pounds apiece, can you buy for 

 7350 pounds ? 



5. How many barrels will it take to pack 36800 pounds of 

 pork, allowing 200 pounds to a barrel ? 



10. We do not go into a detailed explanation of the following 

 tifices, which are often useful in performing calculations with- 

 out writing, or in mental arithmetic, as it is called. The truth 

 f them will readily be seen by any one who has mastered the 

 previous processes, and their explanation will be a useful exer- 

 cise for the student. 



11. To multiply by 5. Annex to the multiplicand, and 

 divide by 2. 



To divide by 5. Multiply by 2 and cut off the last figure, Jialf 

 of which will be the remainder. 



To multiply by 15. Annex 0, and to the result add its half. 

 _ To divide by 15. Multiply by 2, cut off the last figure, and 

 divide by 3 ; prefix the remainder so obtained to the figure cut 

 off ; half the number so formed will be the true remainder. 

 EXAMPLE. To divide 327 by 15 : 

 2 x 327 = 654 

 3)65,4 



21 quotient, 



Leaving 2 as remainder from 65. 



Putting this 2 before the figure cut off viz., the 4 we get 24, 

 which divided by 2 gives 12, the full remainder. 



To multiply by 75. Annex two ciphers to the dividend, and 

 subtract from it its fourth part. 



To divide by 75. Multiply by 4, cut off two figures, and 

 divide by 3. Place before the two figures cut off the remainder 

 got by dividing by 3, and divide the number so obtained by 4 ; 

 this will give the whole remainder. 

 Thus, to divide 2351 by 75, we have 

 2351 



3)94,04 



i 



31 for quotient, 

 With remainder 1 from the 94. 



Prefixing this 1 to the 04 cut off, we have 104, which divided 

 by 4 gives 26, the full remainder. 



To multiply by 125. -Annex three ciphers, and divide by 8. 



To divide by 125. Multiply by 8, and cut off the three right- 

 hand figures. These three figures divided by 8 give the re- 

 mainder, the other figures being the quotient. 



The truth of these processes will be better understood after 

 the learner has read the chapter on Fractions. 



EXEKCISE 14. 



1. Work the following sums in division by means of the 

 artifices shown above : 



1. 6035 -J- 5. 7. 3875 -i- 125. 13. 7853 + 55. 



2. 32561 -4- 5. 8. 1125 V 75. 14. 4860 * 25. 



3. 1256 -5- 15. 9. 3825 4- 225. 15. 94880 H- 25. 



4. 3507 * 45. 10. 8450 -s- 5. 16. 25426 H- 125. 



5. 2350 * 25. 11. 43270 -J- 5. 17. 2876 * 175. 



6. 42340 + 25. 12. 2673 -5- 35. 18. 8250 + 275. 



12. To multiply by a number represented by any number of 

 nines repeated. 



Annex as many ciphers to the multiplicand as there are nines 

 in the multiplier, and from the number so formed subtract the 

 original number. Thus, to find 49276 x 99 



4927600 

 49276 subtract 



4878324 Answer. 

 EXEKCISE 15. 

 1. Work the following examples in multiplication : 



1. 4791 X 99. | 2. 7301 x 999. | 3. 6034 x 999. | 4. 463 X 9999. 

 13. To multiply in one line by a number expressed by two 

 figures. 



To the product of any figure in the multiplicand, multiplied 

 by the units' figure of the multiplier, add the product got by 

 multiplying the figure next on the right of the figure first men- 

 tioned by the figure in the tens' place of the multiplier. Write 

 down the units' figure of the number obtained by this process, 

 and carry on the other (or others) as in common multiplication. 

 EXAMPLE. To multiply 5768 by 73 in one line : 

 5768 

 73 



421064 

 Thus, we say > 



3 _*_ 8 = 24 ; write down 4 and carry 2 



3 x 6 + 2 = 20; 20 + 7 x 8 = 76; write down 6 and carry 7 

 3x7 + 7 = 28; 28 + 7X6= 70; write dowa and carry 7 

 3x5 + 7 = 22; 22 + 7x7 = 71; write down 1 and carry 7 

 7 x 5 + 7 = 42, which write down. 



