0 is 0,* but the 1 I carried makes 1; I set down 1: then, 4 times 3 is 12, the whole of which I set down, because there are no more figures to multiply. So, the work is done, and the whole product is 121024.

The learner will readily perceive that the rule for carrying the tens from the amount in each place to that in the next higher place, is the same in Multiplication as in Addition. I would advise the learner, at first, in multiplying, to set down the amount found in each place, as in Addition; and after having found the total product, to look over the whole work carefully again, in order to correct the errors in it, if there should happen to be any. See the next example.

Several methods of proving Multiplication will be given in Case II.

*If O be multiplied by any number, or any number be multiplied by

O, the product will be 0.

CASE II.

When the multiplier consists of several figures.

RULE.

1. Set down the multiplier under the multiplicand, so that units shall stand under units, tens under tens, &c. as in Addition and Subtraction, and draw a line underneath.

2. Multiply the multiplicand by the right hand figure of the multiplier, and set down the product, as in Case first. 3. Multiply, in like maner, by the next figure in the multiplier, and place the right hand figure of the product exactly under the figure you multiply by, and place the other figures of the product to the left of this, in regular order, under the corresponding figures of the former product. In like manner multiply the multiplicand by each of the remaining figures of the multiplier, and place the right hand figure of each product directly under the figure you multiply by.

4. Draw a line below the several partial products thus found, and add them together, in the order in which they stand, and their sum will be the total product required.*

Note. When there are ciphers between the significant figures in the multiplier; then, in multiplying, omit or neglect those ciphers; but be careful in multiplying by the significant figures, to set down the right hand figure of each partial product exactly below the figure of the multiplier by which it is produced.

5207
245

*Demonstration. After having found the product of the multiplicand by the first figure of the multiplier, as in the former Case, the multiplier is supposed to be divided into parts, and the product is found for the second figure in the same manner: but as this figure stands in the place of tens, the product must be ten times its simple value; and therefore the first or right hand figure of this product must be set in the place of tens; or, which is the same thing, directly under the figure multiplied by. And by proceeding in this manner separately with all the figures of the multiplier, it is evident that we shall multiply all the parts of the multiplicand, by all the parts of the multiplier, or the whole of the multiplicand by the whole of the multiplier; therefore these several products being added-together, will be equal to the whole required product; as in the example annexed.

26035

5 times 5207. 20828 40 times ditto. 10414

=200 times ditto.

1275715-245 times ditto.

EXAMPLES.

1. Multiply 5207 by 245.

Multiplicand, 5207
Multiplier, 245

26035

20828 10414

In performing the work of this example, I first set down the given factors as directed in the Rule, and draw a line below them. Then I multiply the multiplicand by 5, as in Case first, and the product is 26035. I next multiply by 4, and the product is 20828, the right hand figure of which I place under the 4 in the multiplier. I then multiply by 2, and the product is 10414, the right hand figure of which I place under the 2 in the multiplier. Lastly, I draw a line under these partial products, and add them together; and I find their sum to be 1275715; which is the total product required..

Product, 1275715

There are three different methods of proving Simple Multiplication, which are as follows:

First Method.

Multiply the given multiplier by the multiplicand, and the total product thus found, will be the same as that found before, if the work be right.*

* When two numbers are to be multiplied together, either of them may be made the multiplier or multiplicand, and the product will be the same. Hence the reason of this method of proof is obvious.-It may be proper further to inform the learner, that when three or more numbers are

Second method of Proof, by casting out the nines.

1. Cast the 9's out of the sum of the figures in the multiplicand, thus: Add together the figures of the multiplicand; but, in adding, omit all the 9's, and reject 9 from the sum as often as it amounts to 9 or more, always retaining the excess over 9; and the last excess thus found, will be the excess of 9's in the sum of all the figures; which excess set down a little to the right hand of the multiplicand. 2. In like manner find the excess of 9's in the multiplier, and place it against the multiplier.

3. Multiply the excess of 9's in the multiplicand, by the excess in the multiplier, and find, as before, the excess over 9 in the sum of the figures of this product, which excess place against the answer to the question, or the total product of the two given factors, which you wish to prove. Then cast the 9's out of the product last mentioned, as before; and if the excess is the same with that standing against the product, the work may be supposed to be right; but if these excesses differ, the work is erroneous.

*

to be multiplied together, it is not material in what order the several factors are multiplied together: e. g. the product of the numbers 2, 5, and 8, is 80; and if the product of any two of them be multiplied by the remaining number, the final product will be 80; for 2×5×8=80, and 2×8×5=80, and 8x5X2-80.

This rule for proving multiplication depends upon a peculiar property of the number 9, which, except the number 3, belongs to no other digit whatever; viz. that "any number divided by 9, will leave the same remainder as the sum of its figures divided by 9."-The rule may be dem

D

This method of proof is illustrated by the two following

I prove the work of example 5th thus: I begin at the left hand figure of the multiplicand, and say 5 and 3 are 8, and 6 are 14; which sum being more than 9, I cast 9 out of it, and 5 remains: then, (omitting the 9 in the multiplicand,) I say 5 and 3 are 8; which sum, being the excess of 9's in the multiplicand, I set down a little to the right. Then I proceed to the multiplier, and say 3 and 4 are 7; which sum I set down to the right hand of the multiplier. onstrated as follows: Let A and B denote the number of 9's in the two factors to be multiplied, and a and b what remain; then 9A+a and 9B+b will be the factors themselves, and their product is (94×9 B)+(9 A×b)+(9 B×a) +(axb); but the first three of these products are each a precise number of 9's, because their factors are so, either one or both: these therefore being cast away, there remains only axb; and if the 9's also be cast out of this, the excess is the excess of 9's in the total product: but a and b are the excesses in the factors themselves, and axb is their product; therefore the rule is true.

This method of proof, however, is not infallible; for although the work will always be wrong when it proves so according to this rule, yet it may not be right when it proves so; because the right figures may stand in the product, and not stand in the right order; or two or more wrong figures may amount to the same, when added together, as the right ones would. But as the method will usually detect mistakes, and is shorter than the other methods, it is thought useful to be retained.