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more factors, and to multiply first by one and then that product by the other, and so on, as in the following example. Let the two numbers be £4 13s. 3d. and 18.

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Here we first find 9 times the multiplicand, or £41 196. 3d. and then take twice this product, which will evidently be twice 9 or 18 times the original multiplicand (82). Instead of multiplying by 9, we might multiply first by 3 and then that product by 3, which would give the same result; also the multiplier 18 might be resolved into 3 and 6, which would give the same product as the above. If we multiply £83 18s. 6d. by 7,

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we shall have the product of the original multiplicand by 7 times

18 or 126.

If the multiplier were 105, it might be resolved into 7, 3, and 5, and the product be found as above.

But it frequently happens, that the multiplier cannot be resolved in this way into factors. When this is the case, we may take the number nearest to it, which can be so resolved, and find the product of the multiplicand by this number, as already described, and then add or subtract so many times the multiplicand, as this number falls short, or exceeds the given multiplier, and the result will be the product sought. Let there be £1 7s. 8d. to be multiplied by 17.

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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 feet, 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.

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

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, 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 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.

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.↑

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What is the value of 119 yards of cloth at £2 4s. 3d. per Ans. £263 5s. 9d.

yard?

†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.
Sqrs. 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. 0' 10".

13 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′′".

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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.

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we shall evidently come back to the multiplicand, £7 14s. 7d. 3q. We arrive at the same result also, by dividing the above sum re duced, 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.

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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 denomi nation, adding to it the number of this denomination, and divide the sum by the divisor, reducing the remainder, as before, and proceed in

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