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8. How much will 5 bushels of apples come to at 22 cents per bushel ?

Ans. 110 cents. Solution-22 is 2 tens and 2 units, and 5 times 2 units

are ten units or 10; and 5 times 2 tens are 10 tens, and 10 tens and 1 ten make 11 tens=110, the answer.

9. There is an orchard containing 5 rows of trees, in each row are 23 trees; how many trees does the orchard contain ?

By this we see that we may In the first row 23 trees obtain the whole number of trees

second 23 by setting down 23, 5 times and third 23 adding it up, making 115 trees fourth 23

in the whole. fifth 23

But we need only set down

the number of trees contained in In the whole 115 trees 1 row, and as 5 rows will con

tain 5 times as many, we may Multiplicand 23 place the 5 underneath for a mulMultiplier 5 tiplier; we know that 5 times 3

units are equal to 15 units or 1 Product 115 ten and five units; we set down

the 5 units and reserve the 1 ten; then 5 times 2 teñs are 10 tens, and 1 ten which we reseryed makes 11 (tens) which set down, and the answer is 115, as before. Hence we find that Multiplication is a short way of performing Addition.

SIMPLE MULTIPLICATION Teaches to repeat the greater of two simple numbers as many times as there are units in the less or multiplying number, or it is a compendious method of performing many additions.

The number to be multiplied is called the multiplicand. The number you multiply by, is called the multiplier.

The number found by the operation, is called the product.

Both multiplier and multiplicand are called factors.

Note. It will not make any difference in the result of the multiplication, whether we make the greater number the

multiplicand, and the less number the multiplier, or the less number the multiplicand and the greater number the multiplier; for the product of any two factors will be the same in either case: (thus 5 taken 3 times is the same as 3 taken 5 times.) But in practice it is moreconvenient to make the greater number the multiplicand and the less, the multiplier.

CASE I.

When the multiplier is not greater than 12.

RULE.

Place the multiplier under the multiplicand, and multiply, each figure in the multiplicand by the multiplier, setting down and carrying as in Simple Addition.

EXAMPLES.

1. There are 365 days in one year; how many days are there in three years ?

Illustration--It is evident that if one year contains 365 days, three years will contain 3 times 365 days. Operation.

We write down the multiplicand and 365 multiplicand place the multiplier under it

, making 3 multiplier the greater number the multiplicand; 1095 Product the less the multiplier.

We then say 3 times 5 are 15 setting down the 5 (units) and carrying 1 (ten,) as in Simple Addition; then 3 times 6 are 18 and 1 to carry makes 19, setting down 9 and carrying 1 to the next, we say 3 times 3 are 9 and 1 makes 10, setting down the whole product. Hence we find that 3 times 365 is equal to 1095. We might have obtained this same answer by setting down 365 the multiplicand 3 times and adding it up.

3

4. Multiplicand 57436 5432 2345 9054 152634 Multiplier

2
3
4
5

6 Product 114872

2

5

6

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11. If a man travel 48 miles a day, how many miles will he travel in 9 days ?

Ans. 432 miles. 12. What is the product of 956 multiplied by 12 ?

Ans. 11472. 13. What will 325 barrels of flour come to at 7 dollars a barrel ?

Ans. $2275. 14. 320 rods make a mile; how many rods are there in 8 miles ?

Ans. 2560.

CASE II.

When the Multiplier consists of several figures.

RULE.

1. Write the multiplier under the multiplicand, placing units under units, tens under tens, &c.

2. Then multiply by each significant figure in the multiplier, separately, and place the first figure in each product, directly under its multiplier.

3. Add together the several products in the same order as they stand, and you will have the total product.

Proof.*--Multiply the multiplier by the multiplicand.

EXAMPLES.

1. There are 365 days in one year; how many days are there in 25 years ?

Operation. Illustration.—We first multiply Days in 1 year 365 each figure in the multiplicand by Years

25 the 5 (units) in the multiplier, which

gives the product of 5 times 365 or 1825 the number of days in 5 years. 730

We then multiply each figure in

the multiplicand by the 2 tens in the Days in 25 yrs. 9125 multiplier: placing the first figure in

* Another method of Proof is as follows, viz. : cast out the 9's in both factors, and set down the remainder at the right hand ; then if the excess of 9's in the two remainders when muitiplied together be equal to the excess of 9's in the total product, the work is supposed to be right.

Thus, 365=5 excess of 9's.

29-2 excess of 9's.

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the product directly under its multiplier ; and as the value of any number is increased ten times by being moved one place to the left, it is evident that the product found by multiplying by 2 in the ten's place is equal to 20 times 365 or the number of days in 20 years; and the product of 5 years and 20 years added together are equal to 9125 days or the number of days in 25 years.

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37687 When there are ciphers between the

205 significant figures in the multiplier as in 188435

this example; omit the ciphers, and mul75374

tiply by the significant figures only; pla

cing the first figure in each product un7725835 der its multiplier, as before. 8. Multiply 3672 by 508 Product 1865376. 9. Multiply 543764 by 239

129959596. 10. Multiply 181281 by 753

138317403. 11. What number is equal to 3027 times 82164973 ?

Ans. 248713373271. 12. Multiply 27501976 by 271. Ans. 7453035496. 13. Multiply 8496427 by 874359. Ans. 7428927415293. 14. Multiply 95644796 by 8000009.

Ans. 765159228803164 15. Multiply 562916859 by 49007.

Ans. 27586866509013.

CASE III.

When ciphers stand at the right hand of either or both of the factors, neglect them, and place the significant figures under one another and multiply by them only; then place as many ciphers on the right hand of the product, as were neglected in both factors.

EXAMPLES.

Multiply 293

568000

7554000 by 700

84

3400 Product 205100 47712000 25683600000 4. Multiply 4568 by 900.

Ans. 4111200. 5. How many are 800 times 2567 ? Ans. 2053600. 6. Multiply 29526000 by 4030. Ans. 118989780000. 7. Multiply 596780000 by 98200.

Ans. 58603796000000. 8. 930137000 X 9500=8836301500000. 9. 819600000 X 591800000=485039280000000000.

When the Multiplier is 10, 100, 1000, &c. Since numbers increase in a tenfold proportion from the right hand to the left, it is evident that if we place one cipher on the right hand of any figure it increases the value of that figure ten times by removing it to the place of tens ; and if we place two ciphers on the right of any figure it increases its value a hundred fold by removing it to the place of hundreds. Hence the following

RULE.

Place the ciphers of the multiplier on the right of the multiplicand and it makes the product required.

EXAMPLES.

1. Multiply 56 2. Multiply 548 3. Multiply 85 4. Multiply 584 5. Multiply 9640

by 10, the product is 560
by 100

54800
by 1000

85000 by 10000

5840000 by 100000

964000000

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