Rule.—Multiply each remainder by all the divisors preceding the one which obtained it, and take the sum of the products and the remainder arising from the first division. Divide and find the true remainder. 2. 13225 by 105 (3, 5, 7). Ans. 125. Rem. 100. 3. 43125 by 126 (2, 7, 9). Ans. 342. Rem. 33. 4. 141190 by 180 (4, 5, 9). Ans. 784. Rem. 70. 5. 16199 by 216 (2, 3, 4, 9). Ans. 74. Rem. 215. 6. 113546 by 378 (2, 3, 7, 9). Ans. 300. Rem. 146. 7. 363887 by 1512 (2, 3, 4, 7, 9). Ans. 240. Rem. 1007. CASE II. 130. When there are ciphers at the right of the divisor. 1. Divide 9856 by 800. SOLUTION.-8 hundreds are contained in 98 hun- OPERATION. dreds 12 times with a remainder of 200; 800 is not mainder of 200; 800 is not 8100)98156 contained in 56, hence the entire remainder is 200+56, or 256. 12–256 Rule.-I. Cut off the ciphers at the right of the divisor, and as many terms at the right of the dividend. II. Divide the remaining part of the dividend by the remaining part of the divisor. III. Prefix the remainder to the part of the dividend cut off, and the result will be the true remainder. Note.- When the divisor is a unit of any order with ciphers, the remainder will be the figures cut off at the right, and the quotient the figures at the left. Wbat is the value QUO. REM. 2. Of 50483:-700 ? Ans. 72. 3. Of 43802-1500? Ans. 29. 302. 4. Of 723456; 2800 ? Ans. 258. 1056. 5. Of 5793020:13300 ? Ans. 435. 7520. 6. Of 87644300-4570000 ? Ans. 19. 814300. 7. Of 937659000=39800000 ? Ans. 23. 22259000. 83. CASE III. 131. When the remainders are obtained without writing the products and subtracting. 1. Divide 86795 by 37. SOLUTION.—We divide 86 by 37 and find a quo OPERATION. tient of 2; we then multiply 37 by 2, but instead of 37)86795(2345 writing the product and subtracting it from the par 127 tial dividend, we observe what numbers must be 169 added to the product to give the terms of the par 215 tial dividend, and write them for the remainder, 30 Rem. thus: 37 is contained in 86, 2 times; 2 times 7 are 14 and 2 are 16; we write the 2 under the 6; 2 times 3 are 6 and 1 to carry are 7; 7 and 1 are 8; we write the one under the 8, and bringing down 7, the next figure of the dividend, we have 127 for the next dividend ; 37 is contained in 127, 3 times; 3 times 7 are 21, and 6 are 27; hence we write the 6 under the 7; 3 times 3 are 9, and 2 to carry are 11, which increased by 1 make 12; we write thel under the 2, and bringing down, we have 169 for the next dividend, etc. Rule.-I. Obtain the quotient figures in the usual manner. II. Obtain the remainders by observing what number must be added to each partial product to obtain the terms of the partial dividend. III. Bring down the terms of the dividend in the usual manner, and thus proceed until the division is complete. 2. Divide 811332 by 372. Ans. 2181. 3. Divide 1957413 by 453. Ans. 4321. 4. Divide 6419945 by 3007. Ans. 2135. 5. Divide 8074528 by 6328. Ans. 1276. 6. Divide 97547337 by 3891. Rem. 3858. 7. Divide 4223745376 by 180071. Ans. 23456. 8. Divide 170627676887 by 413071. Rem. 25846. CASE IV. 132. When the divisor is a little less than 100, 1000, etc. 1. Divide 7639521 by 96. SOLUTION.-Dividing 7639521 by 100 (or 96 + 4) OPERATION, by cutting off two figures at the right of the dividend, 76395 21 we obtain for the first partial quotient 76395, and a 3056 01 remainder 21. Since the divisor used is 4 more than 122 25 the real divisor, the remainder is too small by 4 times 5 13 76395. Adding 4 times 76395, or 305580, to the re 33 mainder, we find it to be 305580+21=305601, which contains the divisor. Dividing again, we have a quo 79578 %! tient 3056 and a remainder 1. Adding to this remainder 4 times 3056, or 12224, we have a remainder 12225, which still contains the divisor. Dividing again, we have a quotient 122 and remainder 25, to which remainder adding 4 times 122 or 488, we obtain a fourth remainder 513, which being again divided and increased by 4 times 5, gives the true remainder 33. Adding the several partial quotients, and annexing the remainder, we have 7957833, the quotient required. Hence the following Rule.-1. Cut off from the right of the dividend by a vertical line as many terms as there are in the divisor, multiply the part on the left of the line by the difference between the divisor and 100, 1000, etc., and add the product to the number on the right for a true remainder, of which we make a new dividend. II. Divide as before, multiply the new quotient by the difference between the divisor and 100, 1000, etc., add the product to the remainder for a true remainder, and thus proceed until the remainder is less than the given divisor ; the sum of the several quotients with the last remainder, if any, will be the quotient required. Divide the following: 2. 65343214:-999. Ans. 6540855 3. 797876541: 9994. Ans. 79835555 1. 4. 457637892:9998. Ans. 457723435. 5. 6747890343:-99930. Ans. 6752637453. 6. 5434479222:-99800. Ans. 54453588 22 THE PARENTHESIS AND VINCULUM. 133. The Parenthesis, (), denotes that the quantities included are subjected to the same operation. Thus, 18– (9+5) means 18 minus the sum of 9 and 5. 134. The Vinculum, or bar, - is used for the same purpose as the parenthesis, the numbers under it being considered as one quantity. Thus 12—9—3 means that the difference of 9 and 3 is to be subtracted from 12. 1. What is the value of (572–14)-376-35 ? SOLUTION.—572 – 14 equals 558; 376 - 35 equals 341, and 558 - 341 equals 217. Therefore, etc. 2. Of (84793—45832)—(76345–46247)? Ans. 8863. 3. Of (534—46)— 7640—6989+472-12 ? Ans. 297. 4. Of (7000700—2999299)-40040—37737+572 ? Ans. 3999670. 5. Of 8796—2437+210 x (8761-5672+6912) ? Ans. 65696569. 6. Of (656+397)=(247.–166)+25 x 670 ? Ans. 16763. 7. Of 7945—5340 x (549+751)+(5789—5529) ? Ans. 13025. 8. Of (9324+2461–7275) --3471—2432+1216 X (6789— 2507+3364)? Ans. 15292. PRACTICAL EXAMPLES. 1. The product of two numbers is 415638, and one of them is 7697; what is the other ? Ans. 54. 2. The product of three numbers is 2237984, and two of them are 103 and 97; what is the third ? Ans. 224. 3. The dividend is 274500, the quotient 983, and the remainder 243; what is the divisor? Ans. 279. 4. What is the nearest number to 25000 that can be divided by 575 without a remainder ? Ans. 24725. 5. What is the nearest number to 37401 that can be divided by 784 without a remainder? Ans. 37632. 6. Find the value of 29+348 = 6+217X 25—438 : 73 added to 192 • 24+(225–102) x 26. Ans. 8724. 7. A man paid a debt of $105.45 with an equal number of dollars, dimes and cents; how many were there of each kind ? Ans. 95. 8. Find the value of } (9097+6956—2364)-(8765-2721 +2917):6432+5832 X 99—(3278–118503:297— 2790) { +(12965–5273+7391-8771+3349). Ans. 24. 9. The product of three numbers is 196790480, the smallest is 365, and the product of this and the largest is 396755; required the other two factors. Ans. 496; 1087. 10. A New Jersey farmer, wishing to go West, sold bis farm of 150 acres at $84 an acre, and bought prairie land in Illinois for $45 an acre; how many acres did his new farm coutain ? Ans. 280 acres. . 11. A farmer sold an equal number of ducks and turkeys; for the ducks he received $2 each, and for the turkeys $3.50 each; and the whole amount received was $44; how many of each did he sell? Ans. 8. 12. The first edition, 2800 copies, of a book of 480 pages cost me $1543; what did I pay a page for stereotyping, if the press work cost me $125, the paper about 121 cents a copy, and the binding 15 cents a copy? Ans. $1.35. 13. A cistern containing 13500 gal. is filled by two pipes, one discharging 250 gal. an hour and the other 300 gal., but, by a leak in one of the pipes, 100 gal. are lost in an hour; how long will it take to fill the cistern ? Ans. 30 h. 14. Prove and illustrate that the sum or difference of two numbers, divided by any number, will equal the sum or difference of the quotients found by dividing those two numbers by the same number. 15. The day before Christmas, a butcher sold an equal number of ducks and turkeys, and three times as many chickens; he received for the chickens $1.75, for the ducks $2.25, and for the turkeys $4 each, and the whole amount was $92; what was the number of each? Ans. 24; 8; 8. 16. The keeper of a restaurant, counting the currency received from one day's sales, found it to amount to $31.50, one-ninth being in fifty-cent notes, and the rest made up of an equal number of twenty-five-cent and ten-cent notes; how many were there of each? Ans. 7; 80; 80. 17. A farmer's wife took to the store 6 lb. of butter at 45. cents a pound, 4 doz. eggs at 25 cents a dozen, and 2 pair of spring chickens at $1.25 a pair; she received in exchange groceries amounting to $1.75, a pair of scissors at 50 cents, needles and thread at 20 cents, and delaine at 25 cents a yard; how many yards of delaine did she receive? Ans. 15. 18. James Green purchased Norristown Railroad stock to the amount of $5814, and sold part of it for $2756 at $53 a share, losing $4 on a share; but some years after, the road being leased by the Reading Railroad, he sold out at a gain on the whole transaction of $1492; for what did he sell a share? Ans. $91. |