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18 nundreds, with which we unite the 1 hundred reserved, makAndreds. The 5 being written at the left hand of the 2 tens, we

hundreds and 2 tens, or 520 for the number of trees in 20 se being added to 156, the number in 6 rows, we have 676 for umber 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?

We will represent the 3 rows by 3 lines of l's, and the 1, 1, 1, 1, 1, 5 trees in each row by 5 l’s in nach line. Here it is 1, 1, 1, 1, 1, evident that the whole number of l’s are as many times 5 1, 1, 1, 1, 1, as there are lines, or 3 times 5=15, and as many times

3 as there are columns, or 5 times 3=15. 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 prvo 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. Şimple 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 carrying the tens as in Addition. (84) If the multiplier sists of two or more figurés, begin at the right hand and m tiply all the figures of the multiplicand successively by ea fignre of the multiplier, remembering to set the first figure each product directly under the figure by which you an 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. 7. In the division of a prize

14. If a man's income be 1 ampng 207 men, each man's dollar a day, what will be the sare was 584 dollars; what amount of his income in 45 was the value of the prize? years, allowing 365 days to 534 dolls.

each year? Ans. 16425 dolls 207 men

15. A certain brigade con

sists of 32 companies, and 37 38

each company of 86 soldiers; 1068

how many soldiers in the brie gade?

Ans. 2752. Ans. 1105 38 dolls.

16. A man sold 742 thou8. If a man earn 3 dolls. a

sand feet of boards at 18 dol week, how much will he earn

lars a thousand; what did in a year, or 52 weeks they come to? Ang. 156 dolls.

Ans. 13356 dolls. 9. If a man thrash 9 bush- 17. If a man spend 6 cents els of wheat a day, how much a day for cigars, how much Will he thrash in 29 days?

will he spend in a year of 865 Ans. 261 bush. days? Ans. 2190 cts.=$21.90. 10. In a certain orchard

18. If a man drink a glass there are 27 rows of trees,

of spirits 3 times a day, and and 15 trees in each row

each glass cost 6 cents, what bowany trees are there

will be the cost for a year? Ans. 405. Ans. 6570 cts.= $65.70. 11. If a personent 180 in

19. Says Tom to Dick, you a minute, how many will he have 7 times 11 chesnuts, but count in an hour?

I have 7 times as many as you, Ans. 10800. how many have I? Ăns. 539. 12. A man had 2 farms, on

20. In a prize 47 men shared one he raised 360 bushels of equally, and received 25 dob wheat, and on the other 5. lars each ; how large was the times as much; how much prize? Ans. 1175 dolls. did he raise on both?

21. What is the product, Ans. 2160 bush. 308879 by twenty thousand 13. In dividing a certain five hundred 400 three? sum of money among 352,

Ans 6332946137. each man received 17 dollars, 22. What will be the costtor what was the sum divided? 924 tons of potash at 95 dolls Ans. 5984 dolls.

Ans. 87780 dolls. 23. Multiply 848329 by 4009.

Product, 3400950961 24. Multiply 64+70017103--83 by 1876. Prod. 170040 25. 49X15X17X 12X100=how many? Ans. 14994000


a ton?



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 apie 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

of two, or more, other numbers, is called a composile 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 twoz" the last product will be the

2. What is the weight of 82 boxes, 3. Multiply 24 by 36. cach weighing 42 pounds!

Product 89208.
42 6X7
Ans. 3444 lbs. 4. Multiply 8462 by 56.

Product 473872. 91. 6. What will 16 tons of hay cost at 10 dollars a ton?

It has been shown (73) 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 for the answer.

6. A certain army is made up of 125 com consisting of 100 men each; how many men are there in the whole

For the reasons given under example 5, wenner is multiplied 1 100 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 phers 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. 3579000.

Prod, 78910100000. 92. 9. What is the weight of 250 asks of sugár, each weighing 300 log.?

Here 300 may be regarded as a composite number, 25

whose component parts are 100 and 9; 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 figure, 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 ciphors as mare neglected in both factors.



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10. Multiply 3700 by 200.

11. Multiply 7830 by 97000. Prod. 740000.

Prod. 759510000. 13. 12. Peter has 17 chesnuts, and John 9 times as many; how many has John?

170 Here we annex a cipher to 17, which multiplies it by 10. 17 If now we subtract 17 from this product, we have the 17 nine

times repeated, or multiplied by 9. Ans. 153

13. A certain cornfeld contains 228 rowby which are 99 hills long ; how many hills are there? 22800

Annexing two ciphers to 228, multiplies it 100; 228 we then subtract 228 from this product, which

leaves 99 times 223; and in general, Ang. 22572

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 ihus produced, subtract the multiplicand, the remainder will be the answer.

14. Multiply 99 hy 9. . { 15 Multiply 6473 by 999.


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 vany 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 tha: 12 is the difference between 6 and 18. 'tence 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 inuch remains to be paid?

Now to find the difference between 75 and 43, we Operation. write down the 75, calling it the minuend, or numFrom 75 minuend. ber to be diminished, and write under it the 43, 'Take 43 subtrahcnd. calling it the subtrahend, with the units under units

and the tens under tens, and draw a line below, as 32 remainder. 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 7 units of the upper line,

and find the remainder to be 2, which we write bea low 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 an 1 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 remainderHence, to show the correctness of the operation, we have only to recondpore 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?

Here we take 2 from 7, and write the difference, 5, 727 dolls. below the line in the place of units. We now proceed 542 dolls. to the tens, hut find we cannot take 4 tens from 2 tens.

We may, however, separate 7 hundreds into two parts, Ans. 185 dolls. one of which shall be 6 hundred, and the other 1 bundred,

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

Here we cannot take 7 units from 2 units; we must there7 fore take the 1 ten=10 units, with the 2, making 12 units ;

then 7 from 12 leaves 5 for the answer. 5 Ang.

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?

Here we cannot take 8 units from 6 units; we must therefore 406 borrow 10 units from the 400, denoted by the figure 4, which 178 leaves 390. Now joining the ten we borrowed with 6, we have

the minuend, 406, divided into two parts, which are 390 and 16. 228 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 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 fronı 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

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