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vented by PYTHAGORAS, a Greek philosopher, contains all the products which are absolutely necessary to be learned. It is convenient to know the products of numbers still higher.

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In order to find the product of two numbers in this table ; look for one of the numbers in the left hand column, and for the other in the upper row. Then, in a line with the former, and directly under the latter, you will find the product sought. Thus, opposite 9, and under 6, will be found 54, the product of 9 and 6.

The construction of this table is very simple. It may be entirely formed by Addition. In the upper row, let the nine digits be writ. ten down in order. Then, let every figure in this row be added to itself, and the amount placed directly under the figure thus added. By this process, the second row is found. For the third, add each number in the second, to the one above it in the first. For the fourth, add, in the same manner, the third and first; or add the second to itself. For the fifth, add the fourth and first ; or the third and second. For the sixth, add the fifth and first; or the fourth and second; or add the third to itself, and so on.

The former of the two modes of proof, given in the last section, is that found in common Arithmetics. The latter embraces several varieties, which the inquisitive student will discover for himself. The natural mode of proof, by reversing the operation, would be to divide the product by one of the factors. The quotient, of course, ought to be the other factor.

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2516

00

00 Multiplication may also be proved by the casting out of nines, as follows:

Multiply 6726
by 11 | 2

The proof may be explained thus :

672 is even Is and 6. -12

11 is even 9s and 2. 7392 | 3 To multiply 672 by 9 and 2, then, is the same, in effect, as to multiply it by 11. 692X9, is, of course, even 9s. In multiplying 672 by 2, multiply first its even 9s by 2. This product, is, likewise, even 9s. Then, multiply the 6, (excess,) by 2. If this product be even Is, the whole is plainly even 98. if not, the whole will have the same excess, which the product of 6 and 2 has. 6X2=12=9+3. Then, 672X11, ought to have 3 for its excess of 98. Casting out 9s from 7,392, the product obtained above, 3 remains. Hence, we conclude the work right.

From this explanation, we dorive the following rule :
I. REJECT THE 98 FROM THE MULTIPLIER AND MULTIPLICAND.

II. MULTIPLY TIIE EXCESSES TOGETHER, AND REJECT The IS FROM THEIR PRODUCT.

III. REJECT The 9s FROM THE PRODUCT, AND IF THE EXCESS, THUS FOUND, AGREE WITH THIE LAST, THE OPERATION MAY DE CONSIDERED AS CORRECTLY PERFORMED.

We may prove Multiplication in the same manner, by casting out 3s, for the reasons mentioned under Addition.

We have seen that Multiplication is an abbreviation of Addition. It may seem singular that addition should be employed to abbreviate multiplication. It is a fact, however, that by means of what are called Logarithms, the nature of which we cannot explain here, tedious multiplications may be performed by the mere addition of two numbers.

We will here insert some useful hints, not found in the ordinary Arithmetics. They may be classed under the head of

ABBREVIATED MULTIPLICATION. ART. I. It is easier, always, to multiply by small figures, than by large ; because we have smaller numbers to carry. Let us take the following example. 1275

Here, we shall find it most convenient, after obtainiog a product 3825 by 3, to multiply that product by 3 again, instead of awhiplying the 11475

multiplicand by 9. The result will be the saine in either case for

3 times 3 are 9. 118575

93

It will be observed that this operation is the same, in principle, with the rule for composite numbers. Take another example.

8136 Here, after multiplying by 8, we multiply the product thus ob. 248 tained, by 3; which is equivalent to multiplying the multiplicand

by 24, the remaining part of the multiplier. In this case, we obtain 65088 a single product, where the common rule would make two necessary. 195264 Hence, WE MAY OFTEN OBTAIN A PARTIAL PRODUCT, FROM

ONE ALREADY OBTAINED, BY MULTIPLYING ACCORDING TO 2017728 THE RULE FOR COMPOSITE NUMBERE.

When the larger number comes first, we may often find a second partial pro. duct by division.

ART. II. We have seen, ($ xv.) that annexing a cypher to any number, mul. tiplies that number by 10. Now 5 is half of ten; and 6 times any number, is half of 10 times the same number. If, then, we annex a cypher to any number, and then divide that number, with the cypher annexed, by 2; we shall obtain a l'esult, equal to 5 times the original number.

Thus, to multiply 724 by 5; annex first a cypher, thus, 7,240. Then, take half of this, and it is 3,620, which is 5 times 724.

To multiply 531 by 5. Annex a cypher, thus, 5,310. Divide by 2, and the result is 2,655, which is 5 times 531.

Hence, To multiply bij 5, ANNEX A CYPHER TO THE MULTIPLICAND, AND DIVIDE BY 2.

ART. III. 15 is 10 and 5. Therefore, if we multiply a number by 10, and then by 5, and add the two products, their sum will be i5 times the number mul. tiplied. But, ($ xv.) annexing a cypher is multiplying by 10; and (11.) half the product by 10, is the product by 5. Then to multiply 337 by 15.

Multiply first by 10: thus, 3370
Divide by 2: thus,

1685
Add these results: thus, 5055=15 times 337.
Hence, to multiply by 15, ANNEX A CYPHER TO THE MULTIPLICAND, AND
ADD THIS RESULT TO HALF ITSELF.

Art. IV. To multiply by. 100, (i xv.) we annex two cyphers. But 25 is a quarter of a hundred. Therefore, 25 times a number, is a quarter as much as 100 times the same number. Of course, if we multiply by 100, and divide the result by 4, we shall obtain a result equal to 25 times the number multiplied.

Thus, multiply 452 by 25.
It is multiplied by 100 by annexing two cyphers; thus, 45,200
Divide this result by 4, and we obtain

11,300=25times 452. Hence, To multiply by 25, ANNEX TWO CYPHERS TO THE MULTIPLICANI), AND DIVIDE BY 4.

ART. V. 9 is 1 less than 10. Therefore, 9 times a number is once that nuon, her less than 10 times the same number. Now annexing a cypher, ($ xv.) multiplies by 10. If, then, after multiplying by 10, we take away once the number, we have 9 times the number left. Thus, to multiply 87 by 9.

870 First annex a cypher, which multiplies by 10, and then 87 take away once 87.

783=9 times $7. So, because 99 is 1 less than 100, we can multiply by 99, by annexing two cyphers, and subtracting once the multiplicand. And, to multiply by 999, we may annex three cyphers and subtract once the multiplicand, and so on. Hence,

To multiply by 9, or by a number consisting of several 98, ANNEX TO THE MULTIPLICAND AS MANY CYPHERS AS THERE ARE 98 IN THE MULTI. PLIER, AND SUBTRACT ONCE THE MULTIPLICAND. ART. VI. 98 is two less than 100. We may, therefore, "

multiply by 98, by annexing tro cyphers, and subtracting twice the multiplicand. 97 is three less than 100. We may, therefore, multiply by 97, by annexing two cyphers, and subtracting three times the multiplicand. 998 is two less than 1,000. We may, therefore, multiply by 998, by annexing three cyphers, and subtracting twice the multiplicand, and so on. Hence,

When the multiplier consists of 9 or 98, with other figures follow. ing, ANNEX AS MANY CYPHERS AS THERE ARE FIGURES, AND SUBTRACT AS MANY TIMES THE-MULTIPLICAND, AS THE MULTIPLIER FALLS SHORT OF A UNIT OF THE NEXT HIGHER ORDER.

The above abbreviations are the most important of those which may be made in multiplication, without the use of Logarithms, Many others might be given ; but the thorough student will be able to derive them from the rules in the preceding sections, combined with the instances noticed here. Such are rules for multiplying by 19, 29, 39, 49, &c.; and also for multiplying by 55, 155, 95, 995, 955, 555, 50, 550, &c. Indeed, the practised scholar in multiplication will find few cases in which he will not be able to introduce some mode of operation, shorter than that prescribed by the general rule for multiplication, in $ XVII,

SUBTRACTION.

MENTAL EXERCISES.

XX. 1. 3 boys were skating. One fell down. How many did not fall? 1 from 3 leave how many?

2. 5 chairs are standing together, but 2 have broken backs. How many are whole? 2 from 5 leave how many!

3. 6 tumblers stand on å table, and 4 are full of cider. How many are empty ? 4 from 6 leave how many ?

4. 8 bottles are on a shelf, and 3 have their noses broken off. How many are whole ? 3 from 8 leare how many

? 5. George put 9 marbles on a table, and 3 rolled off. How many were left? 3 from 9 leave how many ?

6. 7 lamps stand on a table, and 4 are burning. How many are not burning ? 4 from 7 leave how many ?

7. A boy had 9 quills. He made pens of 4. How many were left? 4 from 9 leave how many?

8. 8 birds were on a tree. A man shot 4. How many were left? 4 from 8 leave how

many

?
9. 7 penknives are on a card. 5 are open.

How many are shut? 5 from 7 leave how

? 10. 6 chiekens are together. 2 are drinking out of a basin. How many are not drinking? 2 from 6 leave how many?

many

11. 12 boys were playing together, and 3 went home. How many were left? 3 from 12 leave how many?

12. A man bought several sheep for 19 dollars. He sold them for 14 dollars. How many dollars. did he lose? 14 from 19 leave how many ?

13. A boy had 16 marbles, and gave away 11. How many had he left ? 11 from 16 leave how many ?

14. In a school, consisting of 20 boys, 13 learn different studies, and the rest learn grammar. How many learn grammar ? 13 from 20 leave how many ?

15. A man bought some goods worth 18 dollars, and to pay for them, gave a barrel of flour worth 8 dollars, and the rest in money. How much money did he give ? 8 from 18 leave how many ?

16. A farmer had 21 sheep, and a dog killed 5. How many had he left ? 5 from 21 leave how many ?

17. In an orchard, consisting of 27 trees, 12 bear apples, and the rest bear pears. How many bear pears ? 12 from 27 leave how many

? 18. George had 26 cents, and spent 11. How many had he left? 11 from 26 leave how many ?

19. Samuel had 12 books, and lost 3. How many had he left ? 3 from 12 leave how many ? 5 from 12 leave how many ? 6 from 12? 8 from 12? 11 from 12? 2 from 12 ? 7 from 12! 10 from 12?

20. 4'from 15 leave how many ? 10 from 15 ? 9 from 15 ? 8 from 15 ! 3 from 15 ? 12 from 15 ? 2 from 15 ? 13 from 15 ? 14 from 15? 7 from 15? 5 from 15? 6 from 15 ? 11 from 15 ?

21. 6 from 17 leave how many ? 5 from 17 ? 7 from 17? 4 from 17 ? 8 from 17 ? 9 from 17? 3 from 17 ? 2 from 17 ? 11 from 17 ? 10 from 17 ? 12 from 17 ?

22. 16 men were in a boat. The boat upset and 12 were drowned. How many were left ? 12 from 16 leave how many

? 5 from 16 ? 4 from 16 ? 3 from 16? 7 from 16 ? 8 from 16 ? 13 from 16 ? 15 from 16? 9 from 16 ? 10 from 16? 6 from 16?

23. 5 from 13 leave how many ? 4 from 13 ? 10 from 13? 7 from 13? 9 from 13 ? 12 from 13? 8 from 13? 6 from 13? 11 from 13?

24. 8 from 14 leave how many? 6 from 14 ? 4 from 14 ? 10 from 14? 7 from 14 ? 2 from 14 ? 12 from 14 ? 11 from 14 ? 3 from 14? 5 from 14? 9 from 14 ?

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