ARITHMETIC. 
II. If the multiplier be a composite num- 
ber, whose component parte do not exceed 
12, multiply first by one of these parts, 
then multiply the product by the other. 
Proceed in the same manner if there be 
more than two. 
£. s. d. 
Ex. l. 256 .. 4 .. 7\ x by 72 = 12 X 6 
12 
3074 .. 15 .. 6 
6 
Ans. 18448 .. 13 .. 0 
£. s. d. 
Ex. 2. 355 .. 13 .. 7| x 180 = 12 X 5 X 3 
12 
4268.. 3. .9 
5 
21340.. 18.. 9 
Ans. 64022 .. 16 .. 3 
The component parts will answer in any 
order ; it is best, however, when it can be 
done, to take them in such order as may 
clear oft some of the lower places in the first 
multiplication, as is done in both the ex- 
amples. The operation may be proved by 
taking the component parts in a different 
order, or by dividing the multiplier in a 
different manner. 
III. If the multiplier be a prime number, 
multiply first by the composite number next 
lower, then by the difference, and add the 
products. 
£. s. d. 
Ex. 576.. 4.. 91X87 = 12X7 + 3 
12 
6914.. 17., 6 
7 
48404.. 2.. 6 
1728 .. 14 .. 41 = 3 times. 
Ans. 50122 ■■ 16 . .lQi 
Here we multiply the given sum by 1 2 
and 7, because 12 X 7 = 84, the answer is 
484041. 2s. 6d. we then multiply the given 
sum by 3, which gives 17281. 14s. 4+ these 
sums added together give the true answer. 
IV. If there be a composite number a 
little larger than the multiplier, we may mul- 
tiply by that number, and by the difference, 
and subtract the second product from the 
first. 
£. s. d. 
Ex. 3976 ..10.. 4i X 34 = 6 X 6 — 2 
6 
23861.. 2.. 3 
6 
143166 .. 13 .. 6 
7953.. 0-9 
Ans. 135213 .. 12 .. 9 
We multiply the given sum by 6 and 6, 
because 6 X 6 = 36 ; the answer is 1431661. 
13s. 6d. we then multiply the sum by 2, and 
subtracting that product from the former 
tve get the true answer. 
V. If the multiplier be large, multiply by 
10, and multiply the product again by 10 ; 
by which means you obtain an hundred 
times the given number. If the multiplier 
exceed 1000, multiply by 10 again, and con- 
tinue it farther if the multiplier require it ; 
then multiply the given number by the unit 
place of the multiplier; the first product by 
ten place, the second product by the hun- 
dred place, and so on. Add the products 
thus obtained together. 
Examples. Multiply 35 1. 14s. 8 id. by 4555. 
£. s. d. £■ s. d. 
35 .. 14 .. 81 X 5 = 178 .. 13 .. 6| = 5 times. 
10 
357 .. 7 .. 1 X 5 = 1786 .. 15 .. 5 = 50 times. 
10 
3575 ..10 ..10 X 5 = 17877 .. 14 .. 2 = 500 times. 
10 
35755.. 8.. 4 X 4 = 143021-13.-4 = 4000 times. 
~ ' Ans. 162864 .. 16 .. 5| = 4555 times. 
