I use the Reduce function solve some problem, such as
Reduce[2k + 1 > 100 && k < 200, k, Integers]//ToString
How can I only get the result like k = 1, k = 2...... instead of the format such as C[1] ∈ Integers && 2831. <= C[1] <= 14999. && k == 1 + 2 C[1].
When I try your example I do not get the same result you say you get.
Try
StringReplace[ToString[Flatten[ReplaceAll[
FindInstance[2k+1>100&&k<200,k,Integers,200],
Rule->Equal]]],"=="->"="]
which instantly returns
{k = 50, k = 51, k = 52, k = 53, k = 54, k = 55, k = 56, k = 57, k = 58,
k = 59, k = 60, k = 61, k = 62, k = 63, k = 64, k = 65, k = 66, k = 67,
k = 68, k = 69, k = 70, k = 71, k = 72, k = 73, k = 74, k = 75, k = 76,
k = 77, k = 78, k = 79, k = 80, k = 81, k = 82, k = 83, k = 84, k = 85,
k = 86, k = 87, k = 88, k = 89, k = 90, k = 91, k = 92, k = 93, k = 94,
k = 95, k = 96, k = 97, k = 98, k = 99, k = 100, k = 101, k = 102, k = 103,
k = 104, k = 105, k = 106, k = 107, k = 108, k = 109, k = 110, k = 111,
k = 112, k = 113, k = 114, k = 115, k = 116, k = 117, k = 118, k = 119,
k = 120, k = 121, k = 122, k = 123, k = 124, k = 125, k = 126, k = 127,
k = 128, k = 129, k = 130, k = 131, k = 132, k = 133, k = 134, k = 135,
k = 136, k = 137, k = 138, k = 139, k = 140, k = 141, k = 142, k = 143,
k = 144, k = 145, k = 146, k = 147, k = 148, k = 149, k = 150, k = 151,
k = 152, k = 153, k = 154, k = 155, k = 156, k = 157, k = 158, k = 159,
k = 160, k = 161, k = 162, k = 163, k = 164, k = 165, k = 166, k = 167,
k = 168, k = 169, k = 170, k = 171, k = 172, k = 173, k = 174, k = 175,
k = 176, k = 177, k = 178, k = 179, k = 180, k = 181, k = 182, k = 183,
k = 184, k = 185, k = 186, k = 187, k = 188, k = 189, k = 190, k = 191,
k = 192, k = 193, k = 194, k = 195, k = 196, k = 197, k = 198, k = 199}
as one long string. See if that works for you. If this isn't what you are looking for then try to explain what you need and I will see if I can fix this for you.
Look up each of those functions in the help system to understand exactly how and why this works
Apply[List, ...] to have it strictly be a list, correct? Or, perhaps it starts as a List, does it not? I like this answer a lot, but I think OP was hoping for the List outcome, even if this method might be better. Oh, hah, you literally just edited it. I like the usefulness of Apply[Or, ...] though, too.
$\endgroup$
Jun 9, 2020 at 2:14
Update: The system option (see Original answer below) can be set temporarily via the Method option of Reduce to get an enumeration of the solutions:
Reduce[2 k + 1 > 100 && k < 200, k, Integers,
Method -> {"DiscreteSolutionBound" -> 150}]
Original answer: Set "DiscreteSolutionBound" larger than the number of solutions:
SetSystemOptions["ReduceOptions" -> {"DiscreteSolutionBound" -> 150}]
Reduce[2 k + 1 > 100 && k < 200, k, Integers]
(* k == 50 || ...|| k == 199 *)
(To reset: SetSystemOptions["ReduceOptions" -> {"DiscreteSolutionBound" -> 10}].)
You can also get the solutions from Solve without set the system option:
sol = Solve[2 k + 1 > 100 && k < 200, k, Integers]
(* {{k -> 50}, ..., {k -> 199}} *)
Update: Solve has a bound, "SolveDiscreteSolutionBound" -> 1000000, which can be set, if needed, via either of the following:
Solve[..., Method -> {"SolveDiscreteSolutionBound" -> n}]
SetSystemOptions["ReduceOptions" -> {"SolveDiscreteSolutionBound" -> n}]
Addendum: To get a string like the OP indicates, one might do the following:
sol /. Rule -> (HoldForm[Set[##]] &) //
Flatten //
ToString //
StringTrim[#, "{" | "}"] &
