Timeline for Prove ${^{b}a} \equiv {^{b+1}} a \pmod {10^{\lfloor{\log_{10} (^{b}a) }\rfloor + 1}} \Rightarrow a=5$ as $a$ and $b$ are two integers greater than $1$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 28 at 2:06 | vote | accept | Marco Ripà | ||
| May 25 at 22:18 | history | edited | John Omielan | CC BY-SA 4.0 |
Actually change the link to the MSE thread to be to my answer, and make it clear I'm the one who wrote it. Also, make a few other changes.
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| May 23 at 23:21 | comment | added | John Omielan | @MaxAlekseyev I assuming you're referring to the first part of my $(1)$. Note the OP defined ${^{b}a}=a^{\left(^{b-1}a \right)}$ if $b\ge 3$. I extended that definition so it's also true for $b=2$, consistent with ${^{2}a}=a^a$. Next, the OP's equivalence relation is ${^{b}a} \equiv {^{b+1}a \pmod {10^{\operatorname{len}({^{b}a})}}}$. The LHS is ${^{b}a}=a^{\left(^{b-1}a \right)}$, and the RHS is ${^{b+1}a}=a^{\left(^{b}a \right)}$, giving the $a^{({^{b-1}}a)} \equiv a^{({^{b}}a)} \pmod{10^{\operatorname{len}({^{b}a})}}$ result I used. If you're referring to something else, please clarify. | |
| May 23 at 23:14 | history | edited | John Omielan | CC BY-SA 4.0 |
Make it clear I'm referring to my specific answer in the linked Math SE thread, add brackets around the exponents (like the OP did) to make this more clear, & make a few other changes.
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| May 23 at 22:31 | comment | added | Max Alekseyev | Accordinh to the question, it should be $a^{{^{b+1}}a} \equiv a^{{^{b}}a} \pmod{10^{\operatorname{len}({^{b}a})}}$. You have $b-1$ in place of $b+1$. The there is a possibility that power of 2 or 5 divides $a^{{^{b}}a}$. | |
| May 23 at 18:23 | comment | added | Marco Ripà | Nice, thank you! | |
| May 23 at 16:19 | history | answered | John Omielan | CC BY-SA 4.0 |