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62. In 1 mile how many b. c. ?

A. 190,080. 63. How many geographical, and statute miles round the earth ?

A. 21,600 geog. mls.-24,912 stat. mls. 64. How many inches in 27 miles ?

A. 1,710,720. 65. How many feet round the earth ? A. 131,535,360. 66. How many inches round the earth? A. 1,578,424,320, 67. How many b. c. round the earth ? · A. 4,735,272,960. 68. How many sq. yds. in 1 sq. ml. 570 acres, and rods?

A. 5,856,642. 69. How many sq. inches on the surface of the earth, there being 198,000,000 square miles ?

A. 794,868,940,800,000,000. 70. In 201,283,905 sq. in. how many acres, &c. ?

A. 32 acres, 14 rds. 8 yds. 1 ft. 5 in. 71. In 224,765 farthings, how many £. &c. A. £234; 2; 7; 1. 72. In 243,648 farthings, how many dollars, at 6s. ? A. $$846.00. 73. In 32 hhds. how many pts. ? How many gi. ? 74. In 17,833 in. how many E. E. 75. Reduce 129,302 sq. in to sq. mls. 76. Reduce 32,257,864,001,000,000 solid inches, to cu. mls. 77. Reduce 5 cubic miles to cu. in. 78. Reduce 27 centuries to seconds. 79. Reduce 37,681,001,002,302,900,000 seconds to centuries. 80. Reduce 879 lb. 9 oz. 19 dwt. 18 grs. to grs 81. Reduce 276,845,300,020,001 grs. to lbs. 82. Reduce 20 T. 19 cwt. 3 qrs. 19. lb. 4 oz, 3 dr. to dr. 83. Reduce 259,030,278,001,005, dr. to lbs. 84. Reduce 274,300,000,000,274,300,000,000 sec. to centuries.

grs. &c.

OBSERVATIONS ON DIVISION, FOR ADVANCED PUPILS.

♡ XXXIV. Some arithmeticians employ a different order of ar. rangement, in preparing numbers for division, from that prescribed by our rule. The latter, however, is the more common and con. venient.

The process of division differs materially from those which have been previously explained, in the manner of commencing. In all other cases we begin on the right, but here we find it necessary to commence on the left. The reason of this is, that remainders are often left in the higher orders, which must, necessarily, be carried down to the lower.

When the divisor is large, we cannot always, at once, determine the quotient figure. When this is the case, it is convenient to di. vide the first figure of the dividend (or the first two if necessary), by the first figure of the divisor. Then, increase this figure of the divi. sor by 1, and divide by it again, in like manner. Two quotient fig. ures will thus, usually, be found, and, from these (with the interme. diate numbers), the true quotient figures may easily be selected. If one figure of the divisor be not sufficient to determine, two, or three may be employed.

Besides the mode of proof, already given, we have the following.

Since the divisor and the integral part of the quotient, multiplied together, reproduce the dividend, with the exception of the remain. der, it is plain that if wo take the remainder from the dividend, and divide the dividend, so diminished, by the quotient, the result will be the divisor. In case the remainder is less than the quotient, we need not subtract it from the dividend, before employing this method ; for, on dividing by the quotient we shall obtain the divisor, with the same remainder; but when the remainder is greater than the quotient, we must subtract it, or we shall obtain a result, greater than the divisor, with a less remainder than before. Hence, it is best always to subtract the remainder first.

Another mode of proof is as follows. We see that in the process of Division, the dividend is exhausted by repeated subtractions, until the remainder left, is too small to contain the divisor. Now these numbers, thus taken from the dividend, together with the final remainder, ought to equal the dividend again. The order in which they stand in the work, shows the order in which they have been taken away, and in which they ought, therefore, to be added. In the following illustration, a star is placed against the numbers to be added. 29)7583(261

58*

178 174*

43
29*

This mode is the same, in principle, with that in @xxxii, for the numbers with the star are the remain. der, and the products of the divisor, by the several quotient figures ; standing only in a different order of arrangement.

14*

7583 From this, we have a mode of proof by rejecting 9s, on the prin. ciple of that in addition. It is as follows:

1. REJECT THE 9S FROM THE REMAINDER, AND FROM EACH NUMBER SUBTRACTED DURING THE DIVISION.

II. REJECT THE 98 FKOM THE SUM OF THE EXCESSES, THUS OBTAINED.

III. REJECT THE 98 FROM THE DIVIDEND, AND IF THEIR EXCESS AGREE WITH THE LAST, THE OPERATION MAY BE CONSIDERED AS CORRECTLY PERFORMED.

Or, as the divisor and quotient, multiplied, make up, (with the remainder,) the dividend, we obtain another rule, from those in Multiplication and Addition combined.

I. REJECT THE 98 FROM THE DIVISOR AND QUOTIENT, MULTIPLY THE EXCESSES THUS OBTAINED, AND REJECT THE Is FROM THEIR PRODUCT.

II. REJECT THE IS FROM THE REMAINDER, ADD THIS EXCESS TO THE LAST, AND REJECT THE IS FROM THE SUM.

III. REJECT THE 9S FROM THE DIVIDEND, AND IF THE EXCESS THUS OBTAINED, AGREE WITH THE LAST, THE OPERATION MAY BE CONSIDERED AS CORRECTLY PERFORMED.

It was mentioned in multiplication, that processes were much shortened by the use of Logarithms. The same may be said of Division. And as multiplication is performed by the addition of Logarithms, so is division, on the other hand, by subtraction.

We will now give a few instances of

ABBREVIATED DIVISION.

Art. I. To divide by 10, we cut off one figure from the right of the dividend. But 5 is half of ten, and therefore 5 is contained in any number as often as 10 is contained in a number twice as great. Thus 5 is contained in 15 as often as 10 is in 30; and 5 is contained in 756 as often as 10 is contained in twice 756. But twice 756 is 1,512: and 10 is contained in 1,512, 151 x times. Therefore, 5 is contained in 756, 151 y times. Hence, to divide by 5,

MULTIPLY THE DIVIDEND BY 2, AND CUT OFF ONE FIGURE FROM THE RIGHT OF THE PRODUCT. THE REMAINING FIGURES WILL BE THE QUOTIENT. THE FIGURE CUT OFF WILL BE SO MANY TENTHS, TO BE ANNEXED TO THE QUOTIENT.

Art. II. To divide by 100, we cut off two cyphers. But 25 is a quarter of 100. 25 will, therefore, be contained in any number as often as 100 is contained in four times that number. Thus, 25 is contained in 100, as often as 100 is contained in 400. And 25 is contained in 1,874, as often as 100 is contained in 1,874X4=7,496. But 100 is contained in 7,496, 74 10 times. Therefore, 25 is contained in 1,874, 74 107 times. Hence, to divide by 25,

MULTIPLY THE DIVIDEND BY 4, AND CUT OFF TWO FIGURES ON THE RIGHT OF THE PRODUCT. THE REMAINING FIGURES WILL BE THE QUO. TIENT ; AND THE FIGURES CUT OFF WILL BE SO MANY HUNDREDTHS, TO BE ANNEXED TO THE QUOTIENT.

ART. III. 9 is 1 less than 10. Therefore, 9 will be contained in any number just as often as 10, and will leave, besides, just as many 18 over. If, then, we tind how many 9s there are in the ones over, and add them to the number of 10s, we shall obtain the whole number of 9s in the given number. For example, in 270 there are 27 tens. Of course, there are 27 nines, and 27"ones over. In these 27 ones are 3 nines. Therefore, in 270, there are 27+3=30 nines.

Again, in 193 are 19 tens and 3 remainder. Therefore, there are 19 nines, and 19 ones orer, and 3 remainder. In the 19 ones are 2 nines and I over. Therefore, in 193 are 19+2=21 nines, and 1 and 3 remainders 21 nines and 4 remainder. Take now an example. Divide 8,875 by 9.. First by 10, thus,

887 5 That quotient by 9, thus 98 Add these quotients and remainders, thus, 985 10 Therefore, there are 985 nines in 8,875, and 10 remainder. But this remainder contains 1 nine and 1 over. Therefore, in 8,875, are 986 nines and I remainder, or 986 Žnines. But this is not all the contraction we can make in this division. The student will at once perceive, that in dividing the quotient 98 by 9, we might employ the artifice used in dividing 8,875. That is, we might first divide it by 10, and then divide the resulting quotient by 9. In this way, we should obtain three quotients and three remainders, to be added together, thus :

5

57719B」!

First divide by 10, thus,

887 This quotient by 10, thus,

88 This quotient by 9, thus,

9 Add these quotients and remainders, thus, 984 But here we obtain a remainder, which contains 2 2 nines. If these be subtracted from the 19, and 2 ad. ded to the quotient 984, we obtain the same result as

986 before, viz. 986 .

But the student will see that we might have used the same artifice in dividing the second quotient, and, for the same reason, a third, fourth, fifth, &c., if we should obtain so many, so that no division by 9 is necessary, at all; thus, First divide by 10, thus,

897

5 This quotient by 10, thus,

88 7 This quotient by 10, thus,

8 This quotient by 10, thus,

0 Then add,

983 | 28
From the remainder 28, take three 9s and add 3 27

them to the 983 nines. Then, as before, we
have,

986 1 Thus, we have made the division by 9 very simple. It can be made more simple still. If in adding the remainders, we had carried from the sum, to the first column in the quotients, exactly as though the numbers were not separated, we should have carried 2, that is, 2 tens, which is 2 units more than 2 nines. We should have, therefore, taken 2 units too much from the remainder. But when we carried, if we had set down the 2, that is, the same figure which we carried, under the remainders' column, to be afterwards united with the remainder, this would have counterbalanced the error. This is done in the following.

887 5 88

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Here we add the column of remainders, exactly as if it was a units' column in Addition. We have 2 to carry. We therefore put a 2 under the 8, and carry the 2 to the first column of quotient figures. Adding the 8 and 2, we have 10, which contains i nine. This 1 nine we subtract from the remainder, and add to the quo. tient; and then we have the same result as before.

985

10

over.

986 We have been thus full upon the division by 9, because it is the same in principle with that which follows, by 99,999, &c. 99'is 1 less than 100. Therefore, 99 will be contained in any number, just as often as 100, with just as many ones

If we find how many 99s there are contained in these ones over, and add them to the number of 100s, it will give us the whole number of 99s in the given number. This case, then, is just like the last, except that we must divide repeatedly by 100, instead of by 9, thus, How often is 99 in 37,645. Divide by 100, thus,

376 | 45 This quotient by 100, thus, 3 76 This quotient by 100, thus, 0 | 03 We observe the same rule 380 | 24 here in carrying as before.

1

380 | 25 The case of 999 is the same, except that we divide repeatedly by 1,000; of 9999, except that we divide by 10,000, &c.

Hence, to divide by a number consisting of 9, or of several 98 :

1. DIVIDE REPEATEDLY BY A UNIT OF THE NEXT HIGHER ORDER, UNTIL YOU OBTAIN 0 FOR A QUOTIENT FIGURE.

II. ADD THE SEVERAL QUOTIENTS, AND LIKEWISE THE SEVERAL REMAINDERS THUS OBTAINED, CARRYING FROM THE REMAINDERS TO THE QUOTIENTS, AS IN COMMON ADDITION, AND PLACING THE FIGURE SO CARRIED, UNDER THE UNITS' COLUMN OF THE REMAINDERS.

III. TO THE TOTAL REMAINDER OBTAINED BY THIS ADDITION, ADD THIS FIGURE, AND IF THE SUM EQUAL OR EXCEED THE GIVEN DIVISOR, SUBTRACT THAT DIVISOR FROM IT, AND ADD 1 TO THE SUM OF THE QUOTIENTS. THE RESULTING QUOTIENT AND REMAINDER WILL BE THE TRUE ONES SOUGHT.

ART. IV. By a similar method we may abbreviate Division, when the units figare of the divisor is not a 9. In order to understand this, we will premise, that the difference between any number and a unit of the next higher order, is called the complement of the number. Thus, 2 is the complement of 98, because 100—98

And 3 is the complement of 97, because 100—.97=3, and so on. Now since 98 is 2 less than 100, 98 will be contained in any number just as often as 100 is contained in the same number, with just as many as over. Then if we find how often 98 is contained in these twos over, and add this number to the number of 100s, we shall obtain the whole number of 98s in the given number. Of course, after dividing by 100, we must multiply the quotient by 2, and divide it by 98. And as we may use the same artifice as before, in this second division, and then in the third, and so on, we find that this case only differs from that in Art. III, in requiring us to multiply by 2 our successive quotients, obtained from repeated divisions by 100.

This will be rendered clear by an illustration.
Divide 75,432 by 98.
Divide by 100, thus,

754 / 32
754X2=1,508. 1503+ 1005

15 | 08 15X2=30. 30-100

0 30

Quo. 769 70 Rem. We will now give a case, in which it will be necessary to carry from the remainders to the quotients. Divide 49,999 by 98. Divide by 100, thus,

499 99 499X2=998. 998-100=

998 9X2=18. 18---100=

. 18

510 | 15

4

Quo. 510 19 Rem. Here, we had 2 to carry to the quotients. As this 2 was 2 hundreds, and as each hundred is 2 more than the divisor, 98, we must increase the remainder by 2X2=4, which is accordingly done above. Instead, then, of adding to the remainder, as before, the figure which we carry, we add that figure, multiplied by the complement 2

It is plain, that if we were dividing by 97, we should be obliged to multiply by 3 wherever, in the above instances, we multiplied by 2. If by 96, we should be obliged to multiply by 4; if by 95, by 5; and so on : that is, we should always multiply by the complement of the divisor. Hence, to divide by a number which consists of 98, except the unit figure,

PROCEED BY THE LAST ROIE, EXCEPT THAT YOU MULTIPLY EACH SUCCESSIVE QUOTIENT BY THE COMPLEMENT OF THE DIVIBOR, BEFORE YOU DIVIDE AGAIN; AND THAT YOU ADD TO THE REMAINDER, THE PRODUCT

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