In the first place, I find the product of £1 7s. 8d. by 16, which is £22 2s. 8d. and to this I add once the multiplicand and this sum £23 10s. 4d. is evidently equal to 17 times the multiplicand.

106. It may be observed, that in those cases, where the decrease of value from one denomination to another is according to the same law throughout, that is, where it takes the same number of a lower denomination to make one of the next higher through all the denominations, the multiplication of one compound number by another may be performed in a manner similar to what takes place with regard to abstract numbers.

This regular gradation is sometimes preserved in the denominations that succeed to fect, in long measure, 1 inch or prime being considered as equal to 12 seconds, and 1 second to 12 thirds, and so on, the several denominations after feet being distinguished by one, two, &c. accents, thus,

10f. 4′ 5′′ 10".

If it were required to find the product of 2f. 4' by 3f. 10', we should proceed as below.

The 4 inches or primes may be considered with reference to the denomination of feet, as 4 twelfths, or, and the 10 inches

20

as, the product of which is, or of, or 40", which reduced gives 3' 4"; putting down the 4", we reserve the 3' to be added to the product of 2 feet by 10', or 1, which product is of a foot, to which 3 being added, we have f. or 1f. and 11'; next multiplying 4' or by 3, we have 1 or 1, which added to the product of 2 by 3 gives 7. Taking the sum of these results, we have 8f. 11' 4", for the product of 2f. 4" by 3f. 10'. The method here pursued may be extended to those cases, where there is a greater number of denominations.

12

Whence, to multiply one number consisting of feet, primes, seconds, &c. by another of the same kind, having placed the several terms of the multiplier under the corresponding ones of the multiplicand, multiply the whole multiplicand by the several terms of the multiplier successively according to the rule of the last article, placing the first term of each of the partial products under its respective multiplier, and find the sum of the several columns, observing to carry one for every twelve in each part of the operation; then the first number on the left will be feet, and the second primes, and the third seconds, and so on regularly to the last.†

The above article relates to what is commonly called duodecimals. The operation is ordinarily performed by beginning with the highest denomination of the multiplier, and disposing of the several products as in the first example below. The result is evidently the same, whichever method is pursued, as may be seen by comparing

What is the value of 9cwt. of cheese at £1 11s. 5d. per cwt? Ans. £14 2s. 9d.

What is the value of 96 quarters of rye at £1 3s. 4d. per quarter? Ans. £112. What is the weight of 7hhds. of sugar, each weighing 9cwt. 3qrs. 12lb? Ans. 69cwt.

In the Lunar circle of 19 years, of 365d. 5h. 48′ 48" each, how many days, &c.? Ans. 6939d. 14h. 27' 12".

Multiply 14f. 9' by 4f. 6'.

Multiply 4f. 7′ 8′′ by 9f. 6'.

Ans. 66f. 4' 6".

Ans. 44f. O' 10".

Required the content of a floor 48f. 6' long and 24f. 3′ broad. Ans. 1176f. 1' 6".

What is the number of square feet &c. in a marble slab, whose length is 5f. 7′ and breadth 1f. 10'? Ans. 10f. 2′ 10′′.

Division of Compound Numbers.

107. A COMPOUND number may be divided by a simple number, by regarding each of the terms of the former, as forming a distinct dividend. If we take the product found in article 105, namely, £63 126s. 63d. 27q. and divide it by the multiplier 9,

this example with that of the same question on the right, performed according to the rule in the text. This last arrangement seems to be preferable, as it is more strictly conformable to what takes place in the multiplication of numbers accompanied by decimals.

we shall evidently come back to the multiplicand, £7 14s. 7d. 3q. We arrive at the same result also, by dividing the above sum reduced, or £69 11s. 9d. 3q. for we obtain one 9th of each of the several parts that compose the number, the sum of which must be one 9th of the whole. But since, in this case, each term of the dividend is not exactly divisible by the divisor, instead of employing a fraction, we reduce what remains, and add it to the next lower denomination, and then divide the sum thus formed, by the divisor. The operation may be seen below.

Whence, to divide a number consisting of different denominations by a simple number, divide the highest term of the compound number by the divisor, reduce the remainder to the next lower denomination, adding to it the number of this denomination, and divide the sum by the divisor, reducing the remainder, as before, and proceed in

this way through all the denominations to the last, the remainder of which, if there be one, must have its quotient represented in the form of a fraction by placing the divisor under it. The sum of the several quotients, thus obtained, will be the whole quotient required.

When the divisor is large and can be resolved into two or more simple factors, we may divide first by one of these factors, and then that quotient by another, and so on, and the last quotient will be the same as that which would have been obtained by using a whole divisor in a single operation. Taking the result of the example in the corresponding case of multiplication, we proceed thus,

By dividing £83 18s. 6d. by 2, we obtain one half of this sum, which, being divided by 9, must give one 9th of one half, or one 18th of the whole. The first operation may be considered as separating the dividend into two equal parts, and the second as distributing each of these into nine equal parts, the number of parts therefore will be 18, and being equal, one of them must be one 18th of the whole.

But when the divisor cannot be thus resolved, the operation must be performed by dividing by the whole at once. If the