Timeline for long enough interval of integers to solve a simultaneous congruence
Current License: CC BY-SA 3.0
4 events
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| Apr 3, 2012 at 23:42 | comment | added | user22202 | Thanks, John, for the clarification. I'm curious what more can be said about the original question. | |
| Apr 3, 2012 at 9:22 | comment | added | Labrador |
Ooops.... got the quantification wrong over the interval, and actually the definitions do match up, once that quantification is inserted. Sorry. (Pointed out to me privately by unknown.) So this relates to the case when the $a_i$'s are distinct primes and $A_i = \{1, a_i-1\}$' for each i, indeed. Thank you.
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| Apr 1, 2012 at 11:03 | comment | added | Labrador | Hi---thanks! It seems that Jacobstahl's function concerns the dual problem, though: "what is a sufficiently short interval length such that for every integer in that interval, there is one $a_i$ such that the reduction mod $a_i$ is not in $A_i$" [for $|A_i| = a_i - 1$]. Whereas the above question is of the type: "What is a sufficiently long interval length such that there exists an integer in that interval such that for every $a_i$ the reduction mod $a_i$ is in $A_i$." The former does give a non-sharp lower bound on latter, but what we want is an upper bound on the latter, no? | |
| Mar 31, 2012 at 4:55 | history | answered | user22202 | CC BY-SA 3.0 |