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We then add these partial products, and obtain 2091 for the total product.

OPERATION.

697 multiplicand. 3 multiplier.

2091 product.

By recalling to mind (ART. 10,) that ten in the place of units are equal to one in the place of tens, ten in the tens' place are equal to one in the hundreds' place, &c. we may perform the above multiplication as follows: First, multiplying 7 of the multiplicand by 3 the multiplier, we obtain 21 units, which are the same as 2 tens and 1 unit. Hence we write down the 1 under the units' column, and reserve the 2 to carry to the tens'. Next, multiplying the 9 by 3, we find 27 tens, to which, adding the 2 tens reserved, we have 29 tens, which are equal to 2 hundreds and 9 tens. Write down the 9 under the tens' column, and reserve the 2 to carry to the hundreds. Finally, multiplying the 6 by 3, we have 18 hundreds; to which add the 2 hundreds reserved, and we have 20 hundreds, the whole of which we write down, obtaining 2093 for the product.

Again, let it be required to multiply 367 by 84. Here the multiplier consists of more than one figure.

Place the multiplier under the multiplicand, units under units, and tens under tens.

Multiplying first by the 4 units, we find 1468 for the product. We are next to multiply by the 8 tens. Now, it is obvious that 1 unit, taken ten times, that is, multiplied by 1

OPERATION.

367 multiplicand 84 multiplier.

1468 2936

30828 product.

ten, must produce 10 units or 1 ten. So 7 units, (as in the example,) multiplied by 8 tens, must produce 56 tens, o 5 hundreds and 6 tens. Therefore, set the first figure, 6 of this second product under the tens' column and reserve the 5 to carry to the hundreds. The next step is the mul tiplication of tens by tens, which must produce hundreds, since 1 ten, taken 1 ten times, is equal to 1 hundred. Therefore 8 tens times 6 tens are 48 hundreds; to which add the 5 hundreds reserved, and we obtain 53 hundreds; equal to 5 thousands and 3 hundreds. Place the 3 under the hurdreds' column, and carry the 5 to the next column, and so proceed throughout. The sum of these partial products will give the total product, 30828.

If the multiplier consists of three figures, its left-hand or hundreds figure, multiplied into the units of the multiplicand, will give hundreds for the first figure of the product, which must of course be set down under the hundreds' column; hundreds and tens, multiplied together, will give thousands; hundreds and hundreds multiplied together will give ten thousands, &c.

If the multiplier consists of four figures, its left-hand or thousands' figure multiplied into units, will give thousands for the first figure of the product, which must be set down under the thousands' column. Thousands multiplied into tens, gives tens of thousands; into hundreds, gives nundreds of thousands; and so on.

It would be necessary to annex ciphers to the figures in these several products, to show their true places, if these places were not determined by the position of the figures with relation to other figures, whose places are known. 16. If we again take the first example, which is to

multiply 697 by 3, we remark that since 697 is to be repeated 3 times, it may be done by writing it down 3 times, and then adding, thus:

697

697

697

2091

And it is obvious that all questions of multiplication may be performed by addition.

Hence, multiplication is sometimes defined as being a concise way of performing several additions.

NOTE.-When a zero or 0 occurs in the multiplier, we may observe that its pro duct must remain 0, since nothing repeated any number of times is still nothing.

PROOF OF MULTIPLICATION.

17. If we interchange the multiplier and multiplicand, and then multiply, we shall obtain the same product if the work is right. (See ART. 15.)

As in addition, these two results may be alike, and still the work may be wrong, since mistakes may occur in both operations. As good proof as any, is to carefully repeat the multiplication.

When 0 is multiplied by any number, what is the result? How is multiplication ometimes defined. How may multiplication be proved? Is this method infallible? Why not? What is as good proof as any other?

CASE I.

18. When the multiplier consists of only one figure.

From what has already been done, we deduce this

RULE.

Place the multiplier under the unit figure of the muitiplicand. Draw a horizontal line underneath.

Then multiply each figure of the multiplicand by the mul tiplier, observing to carry one for every ten, as in addition.

When the multiplier consists of but one figure, how do you proceed? What rule Jo you observe in carrying?

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RULE

I. Place the multiplier under the multiplicand, so that units may stand under units, tens under tens, hundreds under hundreds, &c.

II. Multiply successively by each figure of the multiplier, as in Case I., observing to place the right-hand figure of each partial product directly under the figure multiplied by.

III. Then add together these partial products, and the sum will be the total product sought.

When the multiplier consists of more than one figure, how do you write it? How do you then multiply? How do you add up?

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9. Multiply 3471032 by 70056. Ans. 243166617792.

10. Multiply 1240578 by 302014.

Ans. 374671924092.

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