product 4 is hundreds, with which we unite the hundred reserved, making 5 hundreds. The 5 being written at the left hand of the 2 tens, we have 5 hundreds and 2 tens, or 520 for the number of trees in 20 rows. These being added to 156, the number in 6 rows, we have 676 for the number of trees in 26 rows, or in the whole orchard.

86. 6. There are in a gentleman's garden 3 rows of trees, and 5 trees in each row; how many trees are there in the whole?

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,

We will represent the 3 rows by 3 lines of 1's, and the 5 trees in each row by 5 1's in each line. Here it is evident that the whole number of 1's are as many times 5 as there are lines, or 3 times 5=15, and as many times 3 as there are columns, or 5 times 3-15." This proves that 5 multiplied by 3 gives the same product as 3 multiplied by 5; and the same may be shown of any other two factors. Hence either of the

two factors may be made the multiplicand, or the multiplier, and the pr duct will still be the same. We may therefore prove multiplication by changing the places of the factors, and repeating the operation.

SIMPLE MULTIPLICATION.

87. Simple Multiplication is the method of finding the amount of a given number by repeating it a proposed number of times. There must be two or more numbers given in order to perform the operation. The given numbers, spoken of together, are called factors. Spoken of separately, the number which is repeated, or multiplied, is called the multiplicand; the number by which the multiplicand is repeated, or multiplied, is called the multiplier; and the number produced by the operation is called the product.

RULE.

88. Write the multiplier under the multiplicand, and draw a line below them. If the multiplier consist of a single figure only, begin at the right hand and multiply each figure of the multiplicand by the multiplier, setting down the excesses and carrying the tens as in Addition. (84) If the multiplier consists of two or more figures, begin at the right hand and multiply all the figures of the multiplicand successively by each figure of the multiplier, remembering to set the first figure of each product directly under the figure by which you are multiplying, and the sum of these several products will be the total product, or answer required.(85)

PROOF.

89. Make the former multiplicand the multiplier, and the former multiplier the multiplicand, and proceed as before; if it be right, the product will be the same as the former. (86)

QUESTIONS FOR PRACTICE.

23. Multiply 848329 by 4009.
24. Multiply 64+7001+103-83 by 18+6.
25. 49×15×17×12×100—how many?

CONTRACTIONS OF MULTIPLICATION.

90. 1. A man bought 17 cows for 15 dollars apiece; what did they all cost? If we multiply 17 by 5, we find the cost at 5 dollars apiece, Operation. and since 15 is 3 times 5, the cost, at 15 dollars apiece, will 17 manifestly be 3 times as much as the cost at 5 dollars apiece. 5 If then we multiply the cost at 5 dollars by 3, the product must be the cost at 15 dollars apiece.

85 A number (as 15) which is produced by the multiplication 3 of two, or more, other numbers, is called a composite number. The factors which produce a composite number (as 5 and 3) Ans.$255 are called the component parts.

1. To multiply by a composite number. RULE.-Multiply first by one component part, and that product by the other, and so on, if there be more than two; the last product will be the

answer.

2. What is the weight of 82 boxes,

each weighing 42 pounds?
42 6X7 Ans. 3444 lbs.

3. Multiply 2478 by 36.
Product 89208.

4. Multiply 8462 by 56.

Product 473872.

91. 5. What will 16 tons of hay cost at 10 dollars a ton?

It has been shown (78) that each removal of a figure one place to wards the left increases its value ten times. Hence to multiply by 10, we have only to annex a cipher to the multiplicand, because all the significant figures are thereby removed one place to the left. In the present example we add a cipher to 16, making 160 dollars for the answer. 6. A certain army is made up of 125 companies, consisting of 100 men each; how many men are there in the whole?

For the reasons given under example 5, a number is multiplied by 100 by placing two ciphers on the right of it, for the first cipher multiplies it by 10, and the second multiplies this product by 10, and thus makes it 10 times 10, or 100 times greater; and the same reasoning may be extended to 1 with any number of ciphers annexed. Hence

2. To multiply by 10, 100, 1000, or 1 with any number of ciphers annexed. RULE.-Annex as many ciphers to the multiplicand as there are ciphers in the multiplier, and the number thus produced will be the product. 7. Multiply 3579 by 1000. 8. Multiply 789101 by 100000. Prod. 78910100000.

ibs.?

25
3

Prod. 3579000.

92. 9. What is the weight of 250 casks of sugar, each weighing 300 Here 300 may be regarded as a composite number, whose component parts are 100 and 3; hence to multiply by 300, we have only to multiply by 3 and join two ciphers to the product; and as the operation Ans. 75000 lbs. must always commence with the first significant figare, when the multiplicand is terminated by ciphers, the cipher in that may be omitted in multiplying, and be joined afterwards to the product. Hence

3. When there are ciphers on the right of one or both the factors: RULE. Neglecting the ciphers, multiply the significant figures by the general rule, and place on the right of the product as many ciphers as were neglected in both factors.

Prod. 740000.

10. Multiply 3700 by 200.

11. Multiply 7830 by 97000. Prod. 759510000.

33. 12. Peter has 17 chesnuts, and John 9 times as many; how many

has John? 170

17

Ans. 153

Here we annex a cipher to 17, which multiplies it by 10. If now we subtract 17 from this product, we have the 17 nine times repeated, or multiplied by 9.

13. A certain cornfield contains 228 rows, which are 99 hills long; how many hills are there?

22800
228

Ans. 22572

Annexing two ciphers to 228, multiplies it 100; we then subtract 228 from this product, which leaves 99 times 228; and in general,

4. When the multiplier is 9, 99, or any number of nines. RULE.-Annex as many ciphers to the multiplicand as there are nines in the multiplier, and from the sum thus produced, subtract the multiplicand, the remainder will be the answer.

14. Multiply 99 by 9.

|15 Multiply 6473 by 999.

3. SUBTRACTION.

ANALYSIS.

94. 1. A boy having 18 cents, lost 6 of them; how many had he left? Here is a collection of 18 cents, and we wish to know how many there will be after 6 cents are taken out. The most natural way of doing this, would be to begin with 18, and take out one cent at a time till we have taken 6 cents; thus, 1 from 18 leaves 17, 1 from 17 leaves 16, 1 from 16 leaves 15, 1 from 15 leaves 14, 1 from 14 leaves 13, 1 from 13 leaves 12. We have now taken away 6 ones, or 6 cents, from 18, and have arrived, in the descending series of numbers, at 12; thus discovering that if 6 be taken from 18, there will remain 12, or that 12 is the difference between 6 and 18. Hence Subtraction is the reverse of Addition. When the numbers are small, as in the preceding example, the operation may be performed wholly in the mind;(102) but if they are large, the work is facilitated by writing them down.

95. 2. A person owed 75 dollars, of which he paid 43 dollars; how much remains to be paid?

Operation. From 75 minuend. Take 43 subtrahend.

32 remainder.

Now to find the difference between 75 and 43, we write down the 75, calling it the minuend, or number to be diminished, and write under it the 43, calling it the subtrahend, with the units under units and the tens under tens, and draw a line below, as at the left hand. As 75 is made up of 7 tens and 5 units, and 43 of 4 tens and 3 units, we take the 3 75 proof. units of the lower from the 5 units of the upper line, and find the remainder to be 2, which we write below the line in the place of units. We then take the 4 tens of the lower from the 7 tens of the upper line, and find the remainder to be 3, which we write below the line in the ten's place, and thus we find 32 to be the

difference between 75 and 43. From an inspection of these examples, it will be seen that Subtraction is, in effect, the separating of the minuend into two parts, one of which is the subtrahend, and the other the remainder. Hence, to show the correctness of the operation, we have only to recompose the minuend by adding together the subtrahend and remainder. 96. 3. A person owed 727 dollars, of which he paid 542 dollars; how much remains unpaid?

727 dolls.
542 dolls.

Ans. 185 dolls.

Here we take 2 from 7, and write the difference, 5, below the line in the place of units. We now proceed to the tens, but find we cannot take 4 tens from 2 tens. We may, however, separate 7 hundreds into two parts, one of which shall be 6 hundred, and the other I hundred, or 10 tens, and this 10 we can join with the 2, making 12 tens. From the 12 we now subtract the 4, and write the remainder, 8, at the left hand of the 5, in the ten's place. Proceeding to the hundreds, we must remember that 1 unit of the upper figure of this order has already been borrowed and disposed of; we must therefore call the 7 a 6, and then taking 5 from 6, there will remain 1, which being written down in the place of hundreds, we find that 185 dollars remain unpaid.

4. A boy having 12 chesnuts, gave away 7 of them; how many had he left?

12

7

5 Ans.

Here we cannot take 7 units from 2 units; we must therefore take the 1 ten=10 units, with the 2, making 12 units; then 7 from 12 leaves 5 for the answer.

97. 5. A man has debts due him to the amount of 406 dollars, and he owes 178 dollars; what is the balance in his favour?

406 178

228

Here we cannot take 8 units from 6 units; we must therefore borrow 10 units from the 400, denoted by the figure 4, which leaves 390. Now joining the ten we borrowed with 6, we have the minuend; 406, divided into two parts, which are 390 and 16. Taking 8 from 16, the remainder is 8; and then we have 390, or 39 tens in the upper line, from which to take 170, or 17 tens. Thus the place of the cipher is occupied by a 9, and the significant figure 4 is diminished by 1, making it 3. We then say, 7 from 9 there remains 2, which we write in the place of tens, and proceeding to the next place, say 1 from 3 there remains 2. Thus we find the balance to be 228 dollars.

SIMPLE SUBTRACTION.

98. Simple Subtraction is the taking of one simple number from another, so as to find the difference between them. The greater of the given numbers is called the minuend, the less the subtrahend, and the difference between them the remainder.

RULE.

99. Write the least number under the greater, with units under units, and tens under tens, and so on, and draw a line below. Beginning at the right hand, take each figure of the subtrahend from the figure standing over it in the minuend, and write the remainders in their order below. If the figure