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mainder it will continue to 2(n-r1),
2(2n-r1) &c. until there is no remainder.
Take S6 12b/12c. Divide 12 by 6 =2: no
remainder. Exclusions = 12,
18 24 (=2B.)
Take S6 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
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 S6 10b/50c. First exclusions are
12,[24 &c]; but
20/6=3+2 & 2+4=6 20-12=8 & 2z-8=4
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