figure of the first product which stands in the same place of figures as the figure by which you multiply. Having multiplied by all the figures of the multiplier, add the several products, and the sum is the total product.

117. The reason why the products must be placed as above directed, is as follows: the multiplicand is multiplied by the several figures of the multiplier as though each of these expressed no more than simple units. But any figure towards the left is 10, 100, 1000, &c. times greater than if it expressed only simple units, according to its place; consequently, the product found in multiplying by it, as though it were simple units, should be made 10, 100, 1000, &c. times greater, which is done by removing this product one, two, three, &c. places towards the left. Hence, the first figure of each product must stand as far towards the left as the figure which gave it. To find the product of 6574 × 432, we place the numbers thus:

6574
432

13148 twice the multiplicand.

19722 ten times 3 times, or 30 times. 26296 100 times 4 times, or 400 times.

2839968 product, 432 times 6574,

and, having multiplied by the unit figure 2, as usual, we multiply by the 3 tens in like manner, writing the first figure of this second product in the place of tens-that is, under the tens of the first product; lastly, we multiply by the 4 hundreds, writing the first figure of this product under the place of hundreds in the first product.

As any significant figure is ten times greater when removed one place towards the left, the second product, as it now stands, is ten times greater than if it had stood directly under the first; it is, therefore, 10 times 3 times, or 30 times the multiplicand. Again, the third product, which is 4 times, being removed two places, is 100 times as great as if it had stood directly under the first; it is, therefore, 100 times 4 times, or 400 times the multiplicand. Wherefore, as 400+30 +2 = 432, the sum found by adding the three products just as they stand, is 432 times 6574, as was required.

118. When several numbers have the sign into, between them, we first multiply any two of them: we then multiply

the product by another, and this product again by another, and so on, till all are involved.

1. 7 X 4×2×6=336, which is found by multiplying from left to right, thus: 7 X 428; then, 28 X 256; and, lastly, 56 6336. Or, from right to left, thus: 6 X 212; 12 X 448, and 48 X 7 336, as before. It is, therefore, of no consequence, as to the result, in what order the numbers are multiplied. Also, either operation serves as proof to the other.

Let the numbers in the following examples be multiplied from right to left and from left to right. If the product be the same both ways, the work can scarcely be wrong.

119. When there are ciphers between the figures of the multiplier, we omit them and proceed as usual.

1. 596843 × 50009 = 29847521587

29847521587 sum 50009 times.

Having multiplied by 9, we proceed at once, without regarding the ciphers, to multiply by 5, which is 50000, or 5 x 10000. We, therefore, place the unit figure of the product under the tens of thousands in the first product, by which means this product becomes 50000 times the multiplicand. Wherefore the sum of both products is 50009 times 596843.

Let the following be multiplied both ways, as in Art. 118:

2. 908
3. 59008

4. 70063

908 × 70005 =

5. 60007 × 4072 × 9905 =

6. 87006 800097 × 580067 =

120. If either or both of the factors terminate with ciphers, we cut off these ciphers and multiply the numbers without them after which we place as many ciphers on the right of the product as there were cut off from both factors.

The reason for thus proceeding is obvious, if we consider that, for every cipher omitted on the right of either factor, that factor is (56) ten times less. Hence, the other factor is repeated a number of times which is ten times less than it should be; consequently, the product, which is ten times too small, must be multiplied by ten-that is, (55,) it must have the cipher which was omitted placed on the right. 1. 59400 290=17226000

594

29

5346

1188

17226

Having found the product of 594 X 29 17226, we reason thus:

and 290 2910

(55)

consequently, 59400 X 290594 X 100 X 29 × 10, or 59429100 × 10, or 594 X 29 X 1000; wherefore, the product 17226 must be multiplied by 1000; that is, (55,) we must place 3 ciphers on the right, which is the number of ciphers omitted on the right of the two factors. The true product, therefore, is 17226000.

Let the following multiplications be performed both ways:

2. 320×600×780=32×6×78×10000=149760000

3. 50900X25000890=

4. 5903000×5080×6400=

5. 1976000×400070X70500

6. 8900×305007000×30300=

121. A unit followed by ciphers is evidently less than any

other number containing the same number of figures, and (37) greater than any number containing one figure less. Now, if we multiply a number by a unit followed by ciphers, which is merely to place the ciphers following the unit on the right of the given number, the number of figures in the product will, in this case, be one less than the number in both factors. But the number of figures in the two factors can never exceed the number in the product by more than one figure; for, if it could, then might a number greater than a unit followed by ciphers give a less product, which is impossible.

Again, if we place as many ciphers on the right of the multiplicand as there are figures in the multiplier, we shall have just as many figures in the number thus formed, as there are in both factors. But this number is the product of the multiplicand multiplied (37) by a greater number than the multiplier, and is, therefore, greater than the true product. Wherefore, the product of two numbers cannot contain more figures than are contained in both factors; for, if it could, then might a less multiplier give a greater product, which is impossible.

From the above, therefore, we infer that the product of two numbers contains as many figures as are contained in both, or one figure less.

To predetermine whether there will be as many or one less, which is often useful, multiply the left-hand figure of one factor by that of the other: if the product exceeds 9, there will be as many: if, with what may be to carry from preceding figures, it cannot exceed 9, there will be one less.

122. We have seen (113) that when the sum of the figures of a number is a multiple of 3 or of 9, the number itself is a multiple of the same number. Now, it is evident that when either of two factors is a multiple of 3 or of 9, their product will also be a multiple of the same number. Wherefore, having found that one of the factors is a multiple of 3 or of 9, if we find that the product is not a multiple of the same number, this is a proof that there is an error in the work which must be corrected. If we find that the product is also a multiple, though this is not absolute, it is presumptive evidence that the work is right. The fallible cases are when the error is the number by which we are proving, or its multiple; or, when errors balance each other, neither of which cases is very probable.

123. If we multiply the binomial value of an alt unit (?

by any figure less than 9, we shall have the binomial value of that figure as an alt figure of the same order as the alt unit. But, in multiplying the nines of the alt unit, we still have a multiple of 9, and in multiplying the unit, we have the figure by which we multiply, considered as single units. Wherefore, if all the nines be taken out of the value of an alt figure less than 9, the remainder will always be the figure itself. If, then, we add all the figures of a number together, and take the nines out of the sum, we shall have the remainder that would result in subtracting 9 from the number as often as possible. This is called casting out the nines; in peforming which we omit 9 or the sum of any two figures which make 9.

To cast the nines out of 7468935, we omit 9, also the figures 6 and 3, 5 and 4; we then say, 7 and 8 is 15, from which we cast 9, by saying, 1 and 5 is 6. Hence, 6 is the remainder that would result in subtracting 9 as often as possible from 7468935, which is the same as to divide by 9.

Proof. 7468935-6-7468929, which is a multiple of 9. 124. When neither factor is a multiple of 3 or 9. Το prove the work of a multiplication, we cast the nines out of both factors, and place the two remainders in the opposite angles of a cross in the form of the sign into. We then multiply the two remainders as the sign indicates, cast the nines out of their product, and place the remainder in the upper angle. Lastly, we cast the nines out of the total product, and place the remainder in the lower angle. If the number in the lower angle agrees with that in the upper, the work is supposed to be right.

We first cast the nines out of 6988543, and 7 remains: we then cast the nines out of 58, and 4 remains. These two remainders, 7 and 4, we place in the side angles of the cross. We then multiply them together, and cast the nines out of their product 28, saying: 2 and 8 is 10; 1 and 0 is 1, which we place in the upper angle. Lastly, we cast the nines out