(In reply to
re: flawed program, but some ideas by Charlie)
Wouldn't have gotten here without the groundwork laid by Charlie. From his work, we know that at the very least, 2921 is in S. And therefore every integer > 2921 that ends in 1 or 6 is in S.
Is 3 in S? 3 would be in S if 9 was in S. 9 would be in S if 81 was in S. 81 would be in S if 81^2 = 6561 was in S. But we know that 6561 is in S, so therefore so is 3.
Is 4 in S? 4 would be in S if 16 was in S. 16 would be in S if 256 was in S. 256 would be in S if 65536 was in S. But we know that 65536 is in S, so therefore so is 4.
Etc.
By repeated squaring, every integer that isn't a multiple of 5 eventually turns into a (sufficiently large) number that ends in a 1 or a 6, which means that every integer > 2 that isn't a multiple of 5 is in S. So the integers that aren't in S are 1 and all the multiples of 5.
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Posted by tomarken
on 2024-02-18 14:17:55 |
re(2): flawed program, but some ideas
