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 apiec 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. If then we multiply the cost at 5 dollars by 3, the product must be the cost at 15 dollars apiece.

5

85

3

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

3. Multiply 24 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 (73) that each removal of a figure one place wards the left increases its value ten times. Hence to multiply by 10, we save 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 dol

the answer. nsisting of 100 men

6. A certain army is made up of 125 com each; how many men are there in the whole For the reasons given under example 5, multiplied 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 78910100000.

lbs.?

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 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 ciphers as sore neglected in both factors.

Prod. 740000.

10. Multiply 3700 by 200.

11. Multiply 7830 by 97000. Prod. 759510000.

3. 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 223; 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 rany 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. 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 Inuch 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 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

75 proof.

-

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 recom pose 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.

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
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

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 1 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

in the lower line be greater than the figure standing over it, suppose ten to be added to the upper figure, and the next sig nificant figure in the upper line to be diminished by 1, (96) regarding ciphers, if any come between, as 9s,(97); or, which gives the same result, suppose 10 to be added to the upper figure, and the next figure in the lower line to be increased by 1, with which proceed as before, and so on till the whole is finished.

PROOF.

100. Add together the remainder and the subtrahend, and if the work be right, their sum will equal the minuend.

QUESTIONS FOR PRACTICE.

101. 1. Divide 24 apples equally among 6 boys, how many wil each eceive?

The most simple way of doing this would be, first to give each boy 1 Apple, then each boy 1 apple more, and so on, till the whole were distribated, and the number of 1's, which each received, would denote his share of the apples, which would in this case be 4. Or as it would take 6 apples to give each boy one, each boy's share will evidently contain as many apples as there are sixes in 24. Now this may be ascertained by subtracting 6 from 24, as many times as it can be done, and the number of subtractions will be the number of times 6 is contained in 24; thus, 246=18, 18-6=12, 6—12=6, and 6—6=0. Here we find that by performing 4 subtractions of 6, the 24 is completely exhausted, which shows that 24 contains 6 just 4 times. Now as Subtraction is the reverse of Addition, 94) it is evident that the addition of 4 sixes, (6+6+6+6=24) must recompose the number, which we have separated by the subtraction of 4 sixes. But when the numbers to be added are all equal, Addition becomes Multiplication, (83) and 24 is therefors the product of 4 and 6, (4X6=24). A number to be divided, and which is called a dividend, is then to be regarded as the product of two factors, one of which, called the divisor, is given to find the other, called the quotient; and the inquiry how many times one number is contained in another, as 6 in 24, is the same as how many times the one will make the other, as how many times 6 will make 24, and both must receive the same answer, viz. 4. Hence to prove Division, we multiply the divisor and quotient together, and if the work be right, the product will equal the dividend.

2. How many yards of cloth will 63 dollars buy, at 9 dollars a yard?