1. Multiply 77 by 64.

77

64

EXPLANATION.-Take the product of the units for the two righthand figures of the product. Then increase the tens figure of the 4928 multiplier by 1 and take the product of the tens figures, thus changed, for the left-hand figures of the product.

220. To multiply two numbers of two figures, in each of which the units figure is 5.

1. Multiply 85 by 65.

85

65

5525

required.

EXPLANATION.-Write 25 (the product of the units) as the two right-hand figures of the product. Then take the product of the tens (86) to which add ha. I their sum (of 8 + 6), making 48 +7, or 55. This we prefix to the 25, and the result, 5525, is the product

When the sum of the tens figures is an odd number, in taking one-half of it the remainder is 5 tens, which must be added to the tens figure of the product.

221. To multiply by using the complements of the numbers.

222. The Complement of a number is the difference between the number and the unit of the next higher order.

Thus, the complement of 7 is 3; of 19 is 81; of 96 is 4 of 985 is 15, etc. 1. Multiply 97 by 96.

97 .. 3 96 .. 4 9312

EXPLANATION.—The complement of 97 is 3, and of 96 is 4. The product of the complements is 12, which gives the units and tens figures of the product. For the remaining two figures of the product, subtract either the lower complement from the upper number or the upper complement from the lower number. 4 from 97, or 3 from 96, leaves 93, which we write to the left of the 12, making 9312, the product required. There should be the same number of figures in the product as in both the multiplier and multiplicand.

The method given above is an excellent one when the compliments of the numbers are small, so that the work may be done mentally.

223. To multiply when both numbers are between 100 and 125 or between 1000 and 1025.

1. Multiply 112 by 108.

112

108

EXPLANATION.-This is a modification of the preceding method. We take the product of the excess of 100 in each number (12 X 8) 12096 for the units and tens figures of the product. Then we add the excess of either number over 100 to the other number (112 + 8 or 12), making 120, which prefixed to the 96, gives us 12096, the product required.

108

224. To multiply by a number a little less or a little greater than 100, 1000, etc.

1. Multiply 5432 by 98.

5432

EXPLANATION.-98 times a number is 100 times the number 98 minus 2 times the number. Hence, annex two ciphers to the 543200 number and from the result subtract 2 times the number. 10864 532336

2. Multiply 573 by 103.

57300

EXPLANATION.—In this example, we add 3 times the multipli1719 cand to the multiplicand with the ciphers annexed, because the 59019 multiplier is 3 more than 100.

When the multiplier is a little greater or a little less than 1000, three ciphers should be annexed to the multiplicand (Art. 82.)

3940

1970

5910

EXPLANATION.-By annexing a cipher to the multiplicand, it is multiplied by 10; and 15 is half as much more than 10. Hence, we add to the multiplicand with the cipher annexed its half.

RULE. To multiply by 15, annex a cipher to the multiplicand and add to the result its half. To multiply by 150, annex two ciphers, and by 1500, annex three ciphers to the multiplicand and proceed as before.

226. To multiply by first making convenient changes in the multiplicand and multiplier without affecting the product. 1. Multiply 25 by 16.

50 X 8

=

EXPLANATION.-Multiplying the multiplicand and 400 dividing the multiplier by the same number do not affect the product.

400

Hence, we may mentally take the product of 25 X 4 and of 164, which equal 100 × 4 or 400.

We may also say 25 X 16 50 X 8; or 25 X 16 = 1600 ÷ 4.

The student by exercising his ingenuity may very soon learn to see relations in numbers which will enable him to perform many of the operations mentally.

227. To multiply by any number of 9's.

1. Multiply 763 by 99.

76300

763

EXPLANATION.-By annexing two ciphers to the multiplicand

it is multiplied one time too often. Hence, we subtract the multi75537 plicand (see Art. 217).

RULE.-Annex to the multiplicand as many ciphers as there are nines in the multiplier, and from the result subtract the multiplicand.

228. To multiply when one part of the multiplier is a fac tor of another part.

. Multiply 6374 by 248.

74 EXPLANATION.-8, the number of units, is a factor of 24, which 248 may be regarded as tens. We first multiply by the 8 units. The 24 tens are 3 times as many tens as there are units, hence the product obtained by multiplying by 8 is multiplied by 3. The sum of the partial products is the entire product.

50992 152976

1580752

2. Multiply 3465 by 412.

13860

3465 EXPLANATION.-4, the number of hundreds, is a factor of 12. 412 which may be regarded as units. We first multiply the number by the 4 hundreds, writing the firs figure the product under hundreds. The 12 units are 3 times as many units as there are hundreds, hence the product obtained by multiplying by 4 is multiplied by 3, and the first figure is written in units' place. The sum of the partial products is the entire product.

41580

1427580

229. To find the product of numbers of two or more figures by cross multiplication.

230. Cross Multiplication is a process by which the product only is written, the partial products being combined mentally.

1. Multiply 38 by 27.

38

27 1026

EXPLANATION.-First, multiply units by units; second, tens by units and units by tens; third, tens by tens. Thus, 7 X 8 = 56. The 6 is written in units' place. 5 is carried to 73 and 2 × 8, making 42. The 2 is written in tens' place. 4 is carried to 2 × 3, making 10, which is written in the next two places of the product.

2. Multiply 346 by 53.

346

EXPLANATION.-1st, multiply units by units; 2d, tens by units, 53 and units by tens; 3d, hundreds by units, and tens by tens; 4th, 18338 hundreds by tens. Thus; 3 X 6 = 18. The 8 is written in units' place. 1 is carried to 3 × 4 and 5 × 6, making 43. The 3 is written in tens' place. 4 is carried to 3 × 3 and 5 × 4, making 33. The righthand 3 is written in hundreds' place. The left-hand 3 is carried to 5 X 3, making 18, which is written in the next two places of the product.

3. Multiply 537 by 426.

537 426 228762

EXPLANATION.-6 times 7 are 42. The 2 is written in units' place. 4 is carried to 6 X 3 and 2× 7, making 36. The 6 is written in tens' place. 3 is carried to 6 × 5, 2 × 3 and 4 × 7, making 67. The 7 is written in hundreds' place. 6 is carried to 25 and 4 × 3, making 28. The 8 is written in thousands' place. 2 is carried to 4 X 5, making 22, which is written in the next two places of the product.

If the student will solve the above examples by the ordinary method and then compare the partial products with the above process, he will find that the only difference is that by the latter method we begin to use each figure of the multiplier at the point where we find it begins to affect the final product, while by the common method the entire multiplicand is first multiplied by the units figure of the multiplier, then by the tens figure, etc., and then the partial products are added for the entire product. Careful attention to the order of procedure and a reasonable degree of practice will enable the student to multiply readily by at least two or three places.

7

7

231. To multiply any number containing by itself.

564

1. Multiply 7 by 71.

EXPLANATION.-Multiply the integer by the next higher integer and annex to the product. Thus, 7 X 8 = 56, to which is annexed (the product of the two fractions).

232. To multiply mixed numbers when the integers are alike and the sum of the fractions is 1.

EXPLANATION.-To the product of the integer by the next higher integer annex the product of the fractions. Thus 8 X 9 = 72, to which is annexedor, making 725.

233. To multiply any two mixed numbers having like fractions.

1. Multiply 8 by 73.

EXPLANATION.-Add the product of the fractions, the product of 7 the integers, and of the sum of the integers. Thus, X } = }, 50 which is written under the fractions. Then, 7 X 8 + 3 (7 + 8) +1066, which is written as the integral art of the product, making the entire product 66.

6641

111

81

881

42

2. Multiply 111 by 81.

EXPLANATION.-In this example the sum of the integers is 19, and of 19 is 42. This is placed under the partial product 88%, to which it is added for the complete result.

RULE.-Multiply the integers. Then multiply each integer by the fraction in the other number to its nearest unit, and add the partial products.

TO ESTIMATE THE WEIGHT OF LIVE CATTLE.

235. The weight of cattle by measurement can only be ascertained approximately.

236. To find the superficial feet:

Measure in inches the girth just back of the shoulder-blade and behind the front legs, and the length of the back from the root of the tail to the forepart of the shoulder-blade. Multiply the girth by the length, and divide by 14. The quotient will be the number of superficial feet.

For a girth of from 5 to 7 fee, allow 23 lbs. to the superficial foot.

For a girth of from 7 to 9 feet foot.

low 31 lbs. to the superficial