Timeline for Can the product of two disjoint subsets of numbers like 7, 77, 777, ... be equal?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jan 8 at 7:06 | history | suggested | CommunityBot | CC BY-SA 4.0 |
Correct grammar, especially removing commas before "because", which are grammatical in some languages, but not English
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| Jan 8 at 2:14 | review | Suggested edits | |||
| S Jan 8 at 7:06 | |||||
| Jan 7 at 17:20 | vote | accept | Iulian Serbanoiu | ||
| Jan 8 at 21:18 | |||||
| Jan 7 at 16:12 | comment | added | Aleksei Kulikov | @IulianSerbanoiu if your goal was to get a 6th grade solution, then, more or less by definition, this question would not be acceptable for mathoverflow, in my opinion. | |
| Jan 7 at 16:10 | comment | added | Iulian Serbanoiu | I will try to get from that to a 6th grade level - fingers crossed - I'm starting to dust off my math. | |
| Jan 7 at 16:07 | comment | added | Aleksei Kulikov | @CommandMaster indeed, it is not enough, hence why before posting this I asked OP whether he will be satisfied with the answer to the weaker question. | |
| Jan 7 at 16:06 | comment | added | Daniel Weber | I don't think this is enough for the previous question, on whether the product can be a perfect square, because it's possible that $v_p(10^n - 1) = 2$. Do you know if Zsigmondy's theorem can be strengthened to require $v_p(a^n - b^n) = 1$? | |
| Jan 7 at 16:02 | history | answered | Aleksei Kulikov | CC BY-SA 4.0 |