Subtraction of Compound Numbers.

104. THIS operation is performed in the same way as the subtraction of simple numbers, except with regard to the number which it is necessary to borrow from the higher denominations, in order to perform the partial subtractions, when the lower number exceeds the upper. For instance,

In performing this example, it is necessary to borrow, from the column of shillings, 1 shilling or 12 pence, in order to effect the subtraction of the lower number, 4, and we have for a remainder 8 pence. There now remain in the upper number of the column of shillings only 2; it is necessary therefore to borrow, from that of pounds, 1 pound or 20 shillings, we thus make it 22, of which, when the lower number, 17, is subtracted, 5 remain; we must now proceed to the column of pounds, remembering to count the upper number less by unity, and finish the operation as in the case of simple numbers.

The method of proving subtraction of compound numbers, like that for simple numbers, consists in adding the difference to the less of the two numbers.

Multiplication of Compound Numbers.

105. We have seen, that a number consisting of several denominations may be reduced to a single one, either the lowest or the highest of those contained in it, in which state it admits of being used as an abstract number. But when it is required to find the product of two numbers, one of which only is compound, the sim

plest method is, to consider the multiplication of each denomination of the compound number by the simple factor, as a distinct question, and the several results, thus obtained, will be the total product sought. If it were proposed, for example, to multiply 7. 14s. 7d. 3q. by 9, it may be done thus,

and 631. 126s. 63d. 27q. is evidently 9 times the proposed sum, because it is 9 times each of the parts, which compose this sum. But 27q. is equal to 6d. 3q. and adding the 6d. to the 63d. we have 69d. equal to 5s. 9d. adding the 5s. to the 126s. we obtain 131s. equal to 67. 11s. and lastly, adding the 67. to the 631. we have 697. 11s. 9d. 3q. equal to the above result, and equal to the product of

77. 14s. 7d. 3q. by 9.

Instead of finding the several products first, and then reducing them, we may make the reductions after each multiplication, putting down what remains of this denomination, and carrying forward the quotient, thus obtained, to be united to the next higher product.

Hence, to multiply two numbers together, one of which is compound, make the compound number the multiplicand and the simple number the multiplier, and beginning with the lowest denomination of the multiplicand, multiply it by the multiplier, and divide the product by the number which it takes to make one of the next superior denomination; putting down the remainder, add the quotient to the product of the next denomination by the multiplier, reduce this sum, putting down the remainder and reserving the quotient, as before, and proceed in this manner through all the denominations to the last, which is to be multiplied like a simple number.

When the multiplier exceeds 12, that is, when it is so large that it is inconvenient to multiply by the whole at once, the shortest method is to resolve it, if it can be done, into two or

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.

Here we first find 9 times the multiplicand, or £41 19s. 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,

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. 3d. to be multiplied by 17.