Elementary arithmetic, with brief notices of its history... by Robert Potts.

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Secondly. The divisor 5342 is contained 20 times in the second dividend 128208. When 20 times 5342 is subtracted from this dividend,
the remainder is 21368.
Thirdly. The divisor 5342 is contained 4 times in the third dividend 21368. When 4 times 5342 is subtracted from this dividend, the
remainder is 0.
The process may be thus exhibited:
5342)1730808(300 + 20 + 4
1602600
128208
106840
21368
21368
0
In practice the value of the units and the ciphers are omitted,
and the process is thus performed:First. The divisor 5342 is found to be contained 3 times in the
first five figures of the dividend. The product of 5342 by 3 is subtracted, and to the remainder 1282 is annexed 0, the sixth figure of
the dividend.
Secondly, The divisor 5342 is contained 2 times in 12820, and the
product of 2 times 5342 is subtracted from this number, and to' the
remainder 2136 is annexed 8, the seventh figure of the dividend.
Thirdly. The divisor 5342 is contained 4 times in 21368, and 4
times 5342 subtracted from 21368 leaves no remainder.
And 5342 is contained 324 times exactly in 1730808.
The process is thus shewn:
5342)1730808(324
16026
12820
10684
21368
21368
0
When there are one or more ciphers at the right hand of the
divisor they may be omitted, and as many figures from the right hand
of the dividend, and the division performed with the remaining figures,
and at the end of the process, to the remainder must be annexed
the figures omitted in the dividend, to make up the whole remainder.
10. Any number is divided by 10, 100, 1000, &c., by marking off
one, two, three figures from the right hand of the dividend. These
will be the remainder, and the rest will form the quotient.
Any number can be divided by 9, 99, 999, 9999, &c., by successively
dividing the given number by 10, by 100, by 1000, by 10,000, &c.,
respectively, and taking the sum of the successive quotients for the
true quotient, and the sum of the successive remainders for the true