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## Henry Bushby Transcription pages

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85

mainder it will continue to 2(n-r^{1}),

2(2n-r^{1}) &c. until there is no remainder.

Take S_{6} ^{12b}/_{12c}. Divide 12 by 6 =2: no

remainder. Exclusions = 12, ~~18~~ 24 (=2B.)

Take S_{6} ^{10b}/_{50c}. Divide 10 by 6 =1+4

2B=12. Exclusions are 2n(=12)

It may be said at once that

the first exclusion is 2n, the 2nd

4n, the 3rd 6n, the 4th 8n, but

where ~~these nos. [?]~~ one of these nos.

equals x, there the thing ends, i.e.

if 2B is divisible by 2n (i.e. B by

n). But if it is not then the next

exclusion after the one lower than

2B is the difference from 2B + balance

of 2n, i.e. it is a [sum?] equal to 2n.

Take S_{6} ^{10b}/_{50c}. First exclusions are

12,[24 &c]; but ~~20/6=3+2 & 2+4=6~~ *20-12=8 & 2z-8=4*