Timeline for Square and reversed integer
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jun 18 at 7:52 | comment | added | Jérôme JEAN-CHARLES | In base $b$ , let $m$ be maximum digit of numbers satisfying your $k$ relation. Then $m^k \lt b$. So the general solution is any digit is at most the $k$'th root of $b$. | |
| May 8, 2013 at 19:18 | comment | added | user12806 | Thanks for the answer. But why are the digits lower or equal than $3$ ? | |
| May 7, 2013 at 10:40 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 3.0 |
added 374 characters in body; added 89 characters in body
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| May 7, 2013 at 10:28 | history | answered | Nilotpal Kanti Sinha | CC BY-SA 3.0 |