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Henry Bushby Transcription Pages


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on p269
In the same way it may be shown that
the G.C.M. ^(when there is one) of n & B is the G.C.M. of n & C.
For C = M(n-1)
& B = C[over] n -1 = M(n-1) [over] n-1 = M.
[symbol therefore] G.C.M. of n & B = that of n & M = that
of n & M multiplied by any no. prime to n
e.g. Mx(n-1) which is C.

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