Timeline for Can the product of two disjoint subsets of numbers like 7, 77, 777, ... be equal?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 9 at 21:56 | comment | added | kaya3 | Dare I ask if the problem statement stipulates that $n > 0$? | |
| S Jan 8 at 21:20 | history | suggested | user3840170 | CC BY-SA 4.0 |
unnecessary n-ary operator, tag(?)
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| S Jan 8 at 21:18 | vote | accept | Iulian Serbanoiu | ||
| Jan 8 at 14:59 | review | Suggested edits | |||
| S Jan 8 at 21:20 | |||||
| Jan 7 at 23:59 | history | became hot network question | |||
| Jan 7 at 21:50 | answer | added | Aleksei Kulikov | timeline score: 36 | |
| Jan 7 at 19:36 | history | edited | Iulian Serbanoiu | CC BY-SA 4.0 |
added 8 characters in body
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| Jan 7 at 17:20 | vote | accept | Iulian Serbanoiu | ||
| S Jan 8 at 21:18 | |||||
| Jan 7 at 16:53 | history | edited | YCor |
edited tags
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| Jan 7 at 16:49 | review | Close votes | |||
| Jan 21 at 3:07 | |||||
| Jan 7 at 16:47 | answer | added | KhashF | timeline score: 9 | |
| Jan 7 at 16:10 | comment | added | Marco Ripà | @IulianSerbanoiu I would start from the divisibility criterion by $7$. Then we can observe that $42=6 \cdot 7$ and this adds one more $7$ to the $6$ $7$s case, so we can repeat the idea stated above for the cases when the number of $7$ is a multiple of $3^n$. It wouldn't be so hard to get a proof. | |
| Jan 7 at 16:02 | answer | added | Aleksei Kulikov | timeline score: 10 | |
| Jan 7 at 16:01 | comment | added | Iulian Serbanoiu | @MarcoRipà I've reached the conclusion that numbers like 777777 having 6k digis have (at least) an additional 7 when being decomposed, but I still haven't managed to do anything about it. That's why I was initially asking how to prove that the product of all numbers from set M cannot be a perfect square - my intuition tells me that P(A) cannot be equal to P(B) where P represents the product of A/B elements. | |
| Jan 7 at 15:57 | comment | added | Marco Ripà | Maybe it would be interesting to note that is not true that every element of the set $\{7, 77, 777, 7777, \ldots\}$ is squarefree (i.e., we can take a number formed by justaxposing $7$s $3^n$-times in order to get $3^n$ as a factor). Furthermore, it is possible to get numbers that are divisible by $7$ squared... just take a look at $777777$, which is equal to $3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 37$. | |
| Jan 7 at 15:53 | history | edited | Iulian Serbanoiu | CC BY-SA 4.0 |
deleted 32 characters in body; edited title
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| Jan 7 at 15:32 | history | asked | Iulian Serbanoiu | CC BY-SA 4.0 |