If the figures in the units', tens' and hundreds' places be 8, the number is divisible by 8.

If the sum of all the figures be divisible by 3 or 9, the number is divisible by 3 or 9.

If the figure in the units' place be 5 or 0, the number is divisible by 5.

If the sums of the alternate figures beginning at either end be equal, or one sum exceed the other by 11, or by any multiple of it, the number is divisible by 11.

56. To find the least common multiple of two numbers.

To find the least common multiple of 18 and 30, we observe that

so that the least number which contains them both exactly is evidently 6 × 3 × 5 = 90, or the product of 18 and 30 divided by 6 their greatest common measure: and hence, the following rule.

Rule for finding the least common Multiple.

Divide the product of the two numbers by their greatest common measure, and the quotient will be their least common multiple.

If there be more than two numbers, proceed in the same way with the least common multiple of any two of them and the third: and so on, till they are all taken.

Examples of the Least Common Multiple.

57. Required the least common multiples,

58. In the application of the rule last given, the process will be made easier by multiplying either of the numbers proposed by the quotient arising from the division of the other by their greatest common measure: as the result will evidently be the same by each method.

General proofs of all that has been said here, may be found in the Author's Elements of Algebra.

CHAPTER II.

APPLICATION OF ARITHMETIC TO NUMERICAL MAGNITUDES OF VARIOUS DENOMINATIONS, NOT CONNECTED BY THE BASE OF THE COMMON SYSTEM OF NOTATION.

59. IN the preceding Chapter we have considered only such abstract numbers as are formed by figures whose local values are always regulated by the same fixed number ten, which is called the Base of the Common System of Notation: but the rules hitherto given, are easily extended to concrete magnitudes wherein the local values of the different figures are connected by more numbers than one; as for instance, to Pounds, Shillings, Pence and Farthings, where four farthings are equivalent to one penny, which is the next higher denomination; twelve pence to one shilling, which is the next denomination in order; and twenty shillings to one pound, which is the highest denomination here mentioned; the different numbers 4, 12 and 20 connecting the different denominations, in precisely the same manner as the fixed number 10, was supposed to connect the successive denominations of Integers.

The processes employed in cases of this nature are Reduction, and the fundamental operations then called Compound Addition, Compound Subtraction, Compound Multiplication and Compound Division; each of which will be exemplified in order, and the Tables by means of which they are conducted, will be found at the beginning of the work.

REDUCTION.

60. DEF. Reduction is the conversion or changing of numerical quantities, from one or more denominations to one or more others, such that the real or absolute values shall remain unaltered: and its operations will evidently depend upon the principles already explained.

Ex. To reduce £25. 13s. 6d. into farthings, and conversely.

The correctness of the following operations will be manifest from the explanations annexed to their several steps, which are omitted as unnecessary in practice: Direct Operation.

Here the denominations are separated by a point as (.); and this is necessary to distinguish them from ordinary numbers, which do not require it because their local values are all fixed and certain: and it is moreover evident that each of these operations may be regarded as a proof of the other.

61. The former process is sometimes called a descending and the latter an ascending reduction, and they lead respectively to the following rules.

RULE I. To reduce quantities from higher to lower denominations, multiply the highest denomination by the number which connects it with the next inferior, and to the product add the number of the inferior denomination in the quantity proposed; and repeat this for each succeeding denomination till the required one is obtained.

RULE II. To reduce quantities from lower to higher denominations, divide them by the numbers which connect the different denominations in order, and annex the remainders at each step, so as to retain the denominations of the dividends from which they respectively arise.

Ex. How many half-crowns are equivalent to £253. 9s. 10d.?

Here both the rules are requisite, and we have the following operation:

that is, the proposed sum is equivalent to 2027 halfcrowns with 28d. or 2s. 4d. remaining; and this result is verified by reversing the process: thus,

half c. d. 2027.28.

30

12) 6 0 8 3 8 d.

20) 5069.10 d.

£ 25 3.9.1 0.

Examples for Practice.

(1) Reduce £71. 13s. 6d. into farthings; and verify the result.

Answer: 68810 farthings.

(2) Find the number of farthings in 95 guineas 17s. 9ad and conversely.

it.

:

Answer: 96615 farthings.

(3) Reduce £295. 18s. 3 d. to farthings; and prove

Answer: 284079 farthings.

(4) Find the number of pounds, &c., in 415739 farthings; and prove it.

Answer: £433. 1s. 23d.

(5) Reduce 14cwt. 3qrs. 24lbs. into ounces; and prove it.

Answer: 26816 ounces.

(6) Find the number of ounces in 11cwt. 2qrs. 17lbs. 15oz.; and prove it.

Answer: 20895 ounces.

(7) What number of cwts., &c., are contained in 65437 drams? and verify the result.

Answer: 2cwt. 1qr. 3lbs. 9oz. 13drs.

(8) Reduce 3 tons. 14cwt. 3 qrs. 25lbs. 11 oz. 9drs. into drams; and prove the result.

Answer: 2149817 drams.

(9) Find the number of poles contained in 15mi. 5 fur. 31 po.; and verify the result.

Answer: 5031 poles.

(10) In 1081080 inches, how many miles, &c.,? and prove it.

Answer: 17 miles, 110 yards.

(11) Reduce 304935 feet to miles, &c.,; and the

converse.

Answer: 57 mi. 6fur. 5yds.

(12) What number of inches are equivalent to 512 yds. 2ft. 9 in.? and prove the result.

Answer: 18465 inches.

(13) Reduce 54 yds. 8 ft. 104in., superficial measure, into inches.

Answer: 71240 inches.

(14) What number of superficial yards, &c., are equivalent to 40253798 superficial inches?

Answer: 31060 yds. 38 in.

(15) Find the number of cubic yards, &c., in 141721 cubic inches; and prove it.

Answer: 3yds. 1ft. 25 in.

(16) In 5279 pints, how many gallons, &c.,? and prove the result.

Answer: 659 gals. 3qts. 1 pt.

(17) Required the number of weeks, &c., in 72015 hours; and verify the result.

Answer: 428 wks. 4days. 15 hrs.