View Single Post
 2021-02-24, 02:31 #10 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 9,787 Posts $$n=M_{1277}-1 = 2^{1277}-1-1 = 2^{1277}-2 = 2*(2^{1276}-1)$$. 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