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Old 2021-02-24, 02:31   #10
LaurV
Romulan Interpreter
 
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Jun 2011
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\(n=M_{1277}-1 = 2^{1277}-1-1 = 2^{1277}-2 = 2*(2^{1276}-1)\). That's a string of 1276 of 1, followed by a zero. 1276 is divisible by 2, 4, 11, 29. Therefore (by grouping the ones), n should be divisible by all factors of respective mersenne, i.e. 2 (from the last 0), 3 (from M2), 5 (from M4), 23, 89 (from M11), 233, 1103, 2089 (from M29). You can find plenty of combinations which are not true, actually, almost all where you "cross" them. In fact, you can say that the residue is 1 only when you don't cross them (i.e. 23 and 89 are 1 (mod 11), or 233, 1103 and 2089 are 1 (mod 29)), but that is a known property of mersenne factors

Last fiddled with by LaurV on 2021-02-24 at 02:34
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