Timeline for Square and reversed integer
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2013 at 15:06 | history | edited | user9072 |
edited tags; edited tags
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| May 8, 2013 at 0:04 | comment | added | ARupinski | @Gerry Myerson: thanks for clearing that up. Obviously I read the formulation through too quickly without thinking about what it was asserting. | |
| May 7, 2013 at 10:28 | answer | added | Nilotpal Kanti Sinha | timeline score: 1 | |
| May 7, 2013 at 6:27 | answer | added | Gerry Myerson | timeline score: 2 | |
| May 7, 2013 at 6:17 | comment | added | Gerry Myerson | @ARupinski, OP alleges A implies B, you present a counterexample to B implies A. | |
| May 6, 2013 at 23:28 | comment | added | ARupinski | You claim that this is true for $m < 10^8$, but why does for example $m = 32$ work? I have that $f(32\times 32) = 4201 \neq f(32)\times f(32) = 529$ even though for 32 one has $a_0,a_1\in\{0,1,2,3\}$. Is there an extra assumption that is missing here? | |
| May 6, 2013 at 19:24 | comment | added | user12806 | $f(87 \times 87) \neq f(87) \times f(87)$, so we have "$f(87 \times 87) = f(87) \times f(87) \implies 8,7 \in \{0,1,2,3\}$". | |
| May 6, 2013 at 19:18 | comment | added | Aeryk | $9657 = f(7569) = f(87 \times 87) \ne f(87) \times f(87) = 78 \times 78 = 6084$. | |
| May 6, 2013 at 18:51 | history | asked | user12806 | CC BY-SA 3.0 |