we take two thirds of the multiplicand, by taking one third of it 2 times ; thus, 426 x (426 • 3) x 2 284. If the fraction had been annexed to the multiplicand instead of the multiplier, we then would have found the product of the fraction and multiplier, for a partial product, in like manner as above. RULE. — Multiply the fractional part and the whole number separately, and add the products. EXAMPLES OPERATION. 2. Multiply 915 by 22. Ans. 20496. 3. Multiply 12247 by 18. Ans. 220344. 4. If 694 miles make 1 degree, how many miles are 180 degrees? Ans. 12450 miles. 92. When the multiplier is a convenient part of a number of tens, hundreds, or thousands. NOTE. — The following are some of the convenient parts often occurring as multipliers; 2= of 10; 3= £ of 10; 12. = $ of 100; 163 = 1 of 100; 25 = $ of 100; 33} = $ of 100; 125 = $ of 1000; 1663 = ; of 1000; 250 = of 1000; 333}=$ of 1000. Ex. 1. Multiply 785643 by 25. Ans. 19641075. We multiply by 100, by annex4) 7 8 5 6 4 3 0 0 ing two ciphers to the multipli19 6 410 7 5 Product. cand (Art. 65), and obtain a product 4 times as large as it should be, since 25, the multiplier, is only one fourth part of 100; we therefore take one fourth part of that product, by dividing by 4, for the true product. Upon the same principle, if the multiplier had been 3}, we should have annexed one cipher and divided by 3, or, if the multiplier had been 125, we should have annexed three ciphers and divided by 8. Hence the following RULE. — Multiply by the number of tens, hundreds, or thousands, og which the multiplier is a part, and, of the product thus found, take the same part. EXAMPLES. 2. Multiply 68056 by 121. Ans. 850700. 3. Multiply 17924 by 24. 4. Multiply 192378 by 163. Ans. 3206300. 5. Multiply 12345678 by 125. Ans. 1543209750. 6. How much can be earned in one year of 313 working days, at 3dollars a day? 7. What will a farm containing 534 acres cost, at 33} dollars an acre ? 8. In a certain large field there are 771 rows of corn, each containing 250 hills ; how many hills in all ? Ans. 192750 hills. 9. From a port in Louisiana there were exported, in a given time, 9168 boxes of sugar, averaging 1663 pounds to a box; required the whole number of pounds. Ans. 1528000 pounds. 10. An agent has bought for the army, at different times, in the aggregate, 1993 horses, at an average price of 125 dollars each. How much did he pay for them in all ? Ans. 249125 dollars. 11. How many are 333} times 28044 ? Ans. 9348000. OPERATION. 93. When a part of the multiplier is a factor of another part. Ex. 1. Multiply 3263 by 568. Ans. 1853384. We regard the multiplier 3 2 6 3 Multiplicand. as separated into two parts, 5 68 Multiplier. 56 tens and 8 units, or 560 +8; of which the smaller 2 6104 Product by 8 units. part is evidently a factor of 18 2 7 2 8 Product by 56 tens. the larger, since the 56 tens, 18 5 3 3 84 Product by 568. or 560, is equal to 7 tens X 8. We next multiply by the 8 units, obtaining the product for that part of the multiplier. Now, as this product is the same as that by the factor 8 of the other part of the multiplier, we multiply it by 7 tens, obtaining the product of the multiplicand by 8 x 7 tens or 56 tens. These products of the parts, 560 and 8, added together, give the true product by 568. RULE. — Multiply first by the smaller part of the multiplier; and then that partial product by a factor, or factors, of a larger part; and so on with all the parts. The sum of the several partial products will be the product required. Note. — Care must be taken in writing down the partial products to have the units of the different orders stand in their proper places for adding. EXAMPLES. 2. Multiply 112345678 by 288144486. Ans. 32371787641631508. OPERATION. thousands. 11 2 3 4 5 6 7 8 Multiplicand. 6 7 4 0 7 4 0 68 = Product by 6 units. 5 3 9 2 5 9 2 5 44 = Ist product x 8 tens for product by 48 tens. 161 77 7 7 7 6 3 2 2 product * 3 thousands for product by 144 3 2 3 5 5 5 5 5 2 6 4 3d product x 2 millions for product by 288 3 2 3 7 17 8 7 6 41 6 315 08= Product by 288144486. 3. Multiply 61370913 by 96488. Ans. 5921556653544. 4. Multiply 8649347864 by 1325769612. Ans. 11467042561708308768. millions, OPERATION. 94. When the multiplier is any number of nines. Ex. 1. Multiply 87654 by 999. Ans. 87566346. By annexing three ci8 7 6 5 4 000 = 8 7 6 5 4 X 1000 phers to the multiplicand 8 7 6 5 4 = 8 7 6 5 4 X 1 we take it 1000 times, or 1 time more than is re8 7 5 6 6 3 4 6 8 7 6 5 4 X 999 quired by the given mul tiplier. We therefore from this result subtract the multiplicand taken once, and thus obtain the product by 999. In like manner we may multiply by any number of nines. Hence the RULE. — Annex as many ciphers to the multiplicand as there are nines in the multiplier, and from the number thus produced subtract the given multiplicand. NOTE. — To multiply by any number of threes, find the product for the same number of nines, by the rule, and take one third of it by dividing by 3; and to multiply by any number of sixes, take twice the product of the same number of threes, by multiplying it by 2. 2. Multiply 7777777 by 9999. Ans. 77769992223. 3. Multiply 416231 by 99999. 4. Multiply 987654 by 333333. Ans. 329217670782. 5. Multiply 876543 by 66666. Ans. 58435615638. 6. Multiply 999999 by 9999. Ans. 9998990001. 7. Multiply 32567895 by 3333. 8. Multiply 66666 by 66666. Ans. 4444355556. 9. Multiply 912345678 by 99. Ans. 90322222122. 10. Multiply 1234567 by 9999. Ans. 12344435433. 11. Multiply 98123452 by 999999. Ans. 98123353876548. CONTRACTIONS IN DIVISION. OPERATION 95. When the divisor is a convenient part of a number of tens, hundreds, or thousands. Ex. 1. Divide 19641075 by 25. Ans. 785643. By multiplying both divisor and 19 6 410 7 5 dividend by 4, which does not change 4 the relation of the one to the other (Art. 83), the divisor becomes 100, 78 5 6 4 3100 Quotient. which enables us to perform the di vision by simply cutting off two figures at the right of the dividend (Art. 79). In like manner, if the divisor had been 3}, by taking both it and the dividend 3 times as large, we could have performed the division by simply cutting off one figure at the right of the dividend; or, if the divisor had been 125, by taking both it and the dividend 8 times as large, we could have performed the division by simply cutting off three figures at the right of the dividend. Hence the RULE. — Multiply both divisor and dividend by that number which will change the divisor to a number of tens, hundreds, or thousands, and then divide. EXAMPLES. 2. Divide 89630 by 31. Ans. 26889. 3. Divide 123450 by 16. 4. Divide 18621 by 123. Ans. 14899 5. Divide 317121 by 21. Ans. 12684816 6. Divide 876735 by 33}. Ans. 2630216o. 7. Divide 123456 by 125. 8. Divide 61678500 by 250. Ans. 246714. 9. J. Cushing bought a number of horses, at 166 dollars each, for $ 9500; required the number bought. Ans. 57 horses. 10. A company has received 12000 dollars from the sale of piano-fortes, at an average price of 333 dollars each; required the number sold. 11. How many shares of the Illinois Central Railroad, at 125 dollars each, can be bought for 150000 dollars ? Ans. 1200 shares. 12. How many cows, at 33} dollars each, may be purchased for 333} dollars. OPERATION. 13. How many books, at 21 dollars each, may be purchased for 120 dollars ? Ans. 48 books. 14. A certain magazine contains 616350 pounds of powder, in kegs of 25 pounds each ; required the number of kegs. Ans. 24654 kegs. 96.° When the divisor is any number of nines. Ex. 1. Divide 316234 by 99. Ans. 319435. We first divide by 100, or 99 + 1, by cut31 6 2 3 4 ting off two figures at the right of the dividend, 319 6 obtaining for the first partial quotient 3162 127 and a remainder 34. Since the divisor used 28 was 1 larger than the given divisor, 99, the quotient obtained denotes that an excess of 31 9 4 Ans. 3162, or a number equal to the quotient itself, must be added to the 34 for the true remainder, 34 + 3162 3196, which exceeding the divisor 99, we write it for a second dividend; and dividing by 100, or 99 + 1, as before, we obtain for the second partial quotient 31, and a remainder 96. To the remainder 96 we add 31, the excess denoted by the last quotient, and obtain 127 for a third dividend, which being divided by 100, or 99 + 1 gives the third partial quotient 1, and a remainder 27. To the remainder 27 adding 1, the excess denoted by the last quotient, we have for the true remainder 28, which is z. The sum of the partial quotients with the final remainder annexed gives 31947), the quotient required. The above process is based upon the principle that 10 9 +1; 100 = 99 +1; 1000 999 +. 1, etc. ; consequently 20 (2 X 9) + 2, and 37 (3 X 9) + 3 +7; 200 (2 X 99) + 2; 3859 (38 X 99) + 38 + 59 ; 15987 (15 X 999) + 15 + 987, etc. Hence, 316234 316200 + 34 (3162 X 99) + 3162 + 34 ; 3162 + 34 3196 3100 + 96 (31 x 99) +31 + 96; 31 + 96 127 100 + 27 (1 X 99) +1+ 27; 1 + 27 and 3162 +- 31 +1+3= 31947), the answer, as before obtained. By like process, and upon the same principle, may the quotient be found for any number of nines. RULE. — Add 1 to the given divisor, and, by it thus increased, divide by cutting of figures at the right of the dividend. To the figures cut off on the right add those on the left for a true remainder, of which, if it equal or exceed the given divisor, make a second dividend, and divide as before. Proceed thus till there shall be no remainder as large as the given divisor; and the sum of the several quotients, with the last remainder, if any, will be the answer required. NOTE. — When the last remainder is the same as the given divisor, it must be cancelled, and 1 written as a partial quotient. 28 28. |