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In the example, No. 1, say, 7 from 6 I cannot, but 7 from 16 (for you must always borrow 10, in simple numbers, in such cases) and there remains 9, which you set down. Then I that I borrowed and 6 are 7; 7 from 7 and there remains 0; 3 from 8 and there remains 5; nothing from 9 and there remains 9. In order to prove it, work the two : last sums by addition thus:

367 9509

9876 The other súms' are worked and proved in the samą man, ner, taking care, in compound numbers, to attend to the difference of the value of the figures.

Thus, in No. 4, say 9 from7 I cannot, but 9 from 19 (adding 12) and ihere remains 10: 1 that I borrowed and 12 are 13; 13 from 6 I cannot, but 13 froin 26 (adding 20) and there remains 13: 1 that I borrowed and 5 are 6; 6 from 3 I cannot, but 6 from 13 (adding 10) and there remains 7: 1 that I borrowed and 2 are 3 ; 3 from 2 I cannot, but 3 from 12, and there remains 9: i that I borrowed from 4, and there remains S. Prove it as before by addition, thus:

£. d.
25

12 9
S97 13 10
£423

6 7

$.

MULTIPLICATION. This, for general purposes, is the most useful rule in Arithmetic; and therefore particular attention should be paid to the following Table, which must be learned completely by heart before any thing can be done by the pupil to advantage.

In order to understand this table the learner must multiply each figure of the first column by those of the upper row, looking for the product in that square which is in a line with ihe one and underneath the other. Thus if the pupil wants to find the value of 6, multiplied by 5, by looking on the line where the 5 is placed in the first column, under the 6 in the top line, the product will be found to be 30. The way therefore to Icarn this table, which must be

done correctly, is to go on thus; twice 1 are 2, twice 3 are 6, twice 4 are 8, twice 5 are 10, and so on through the whole,

The Multiplication Table.

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3

18

30

33

36

32

36

40

44

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5

30 35

40

45

50

55

60

48

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6 9 12 15 91 24 27 4 8 12 16 20 24 28

10 15 20 25 6 12 18 24 30

30 36 42 7 14 21 8 16 24 32 40 9 18 27 36 45 51 63 72 81 4.90 10 20 30 40 50 60 70 50 90 100 110 120 11 22 33 44 55 66 | 77

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When quite perfect in this, the learner may proceed to the following lesson. EXAMPLES.

d. No. l. Multiply 365 No. 3. Multiply 24 3 6;

6 by

5
1825

£ 145
£.

d.
No.2. Multiply 5420 No. 4. Multiply 452 6 7
by
24

12 21689

£5407 19 0 10840 130,080

1

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In the first of the foregoing exainples say, 5 times 5 are 23 ; then write 5 and carry 2 to the next, saying 5 times 6 are 30, and 2 are 32; write 2 and carry 3, saying 5 tiines 3 are 15 and 3 are 18; therefore 365 multiplied by 5 inakes 1825.

In the second example the multiplier being two figures, namely 24, begin with the 4 and go through the whole of the sum to be multiplied, as in No. 1. Then with the 2 in like manner, only observing to put the product of the first figure under the multiplying figure, as set down, and multiply on as before; when both are performed, the rule of additiou must be applied to ascertain the whole product, as both are to be added, and the amount will be the sum required.

Another way of working this, and which will also prove whether the sum bere stated is right, is, by multiplying by O and 12, because twice 12 are 94, thus:

Multiply 5420

2 10840

12 130080

by

To show, in the third example, the amount of 6 times £24 35.6 d. multiply thus: 6 times 9 are 12; 12 farthings being 3 pence, carry S to the pence: theu 6 times 6 are 56 and 3 are 39; 39 pence beịng 3 shillings and 3 pence, set down 3 and carry 3; then 6 tines S are 18 and 3 are 21; 21 shillings being ! pound 1 shilling, set down ļ and carry 1: then 6 times 4 are 24 and 1 are 25; 5 and carry 2: then 6 times 2 are 12 and 2 are 14. Hence 6 pounds weight, or 6 barrels, or 6 pieces of any article, at 24 $s. 64d. would amount to £ 145 ls. 3d.

When either the multiplier or the multiplicand (that is the sum multiplied) or boil, contains ciphers on the right hand, set down so many ciphers as there are in both, on the right of the products and multiply only by the res mainder, thus: 2405

876500 100

24300 240500

26295 35060 17530 2129895000

DIVISION. As Multiplication teaches the art of finding any number when repeated so many times, so Division instructs us how often one given number is contained in another. Thus, to know how many times 6 are contained in 478654, set them dowu in this manner;

Divide by

6)478654

7977547

This is performed by saying 6's in 47, 7 times and 5 over, because 7 times 6 are 42; then placing the 5 before the next figure 8, it makes 58; 6's in 58, 9 times and 4 over; which placed before the next figure 6 makes 46; then 6's in 46, 7 times and 4 over; 6's in 45, 7 times and 3 over 3 6's in 34, 5 times and 4 over; therefore 79775 and over, is the answer. In order to prove it multiply it thus:

Multiply 79775—4 the Quotient.
by

6 the Divisor.
478654 the Dividend.

Here 6 times 5 are 30, and 4 the remainder are 34; 4 and carry 3, and so on.

When the divisor exceeds 19, it is necessary to proceed as in the following example:

Finding that 10 twenty fives make 25)265(10 250; place 10 on the right and mul

250 tiply the 25 by 10, as here stated, then by subtractiog 250 froin 005, there remains 15 ; so that the answer is 10 times and 15 orer; and in order to prove it, multiply it as before.

15

REDUCTIOX. The next step is to reduce sums of money, &c. into an amount of different denomivations; as, for instance, pounds into shillings, pence, or farthings; years into days, hours, or minutes, &c. It is not, properly speaking, a distinct rule in arithmetic, but rather the application of the two preceding ones, namely, Multiplication and Division,

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S.

In 32

No. 1.
EXAMPLES.

"No. 2.
Of money ascending.

Of money descending. £. d.

14 6: how many farthings. 20

Proved thus : 654 shillings

4)31419 farthings 12

12)78544 7854 pence

2,0)65,4-6d.

32-14s. 31419 answer.

£.32 14 63 answer.

In the first of these examples, begin to multiply by, 20, because 20 shillings make one pound; but as it contains a cipher on the right hand, take the 4 from the 14 shillings, and set down in its proper place; then multiply by the , saying twice 2 are 4 and 1 from the 14 wbich was left are 5, and twice 3 are 6: then multiply the 654 shillings by 12, because 12 pence make one shilling, adding the 6 from the pence to the first figure múltiplied; and lastly multiply the 7854 pence by 4, because 4 farthings make one penny, adding the 3 farthings to the first figure multiplied.

In the second example, the sum is proved by divison, which is the way to ascertain any similar som, here you begin by dividing the 31419 farthings by 4, in order to bring them into pence, thus 4's in 31, 7 times and 3 over; 4's in 34, 8 times, and 2 over; 4's in 21, 5 times, and i over; 4's: in 19, 4 times and 3 over, which are 3-4ths of a penny, and therefore you find that in 31419 farthings, are contained 7854 pence and 3 farthings; thus you proceed through the whole, dividing the pence by 12, because 12 pencemake one shilling, and the shillings by 20, taking care to carry out as here stated the overplus that remains, which must be brought down when the answer is given.

In dividing by 20, cut off the cipher, and the last figure in the quotient, which you carry out ; and divide by 2, it being more easily done; thus 2's in 6, 3 times; 2's in 5, iwice and 1 over, which by carrying out to the 4, makes 14 shillings over.

These two plain examples will give the learner a sufficient idea of the general principle of reduction, as the same method is adopted in the reducing of weights, measures, &c.

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