Intereting Posts

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Confusing myself here, need some clarification..

First, we consider the palindrome $abccba$. We can see this can be written as

$a(10^5 + 10^0) + b(10^4 + 10^1) + c(10^3 + 10^2) = a(10^5 + 1) + 10b(10^3 + 1) + 100c(10 + 1)$. So we see essentially all palindromes of even digits can be written in the form $x(10^{2k+1} + 1)$, so essentially we must show that any number of the form $(10^{2k+1} + 1)$ is divisible by $11$.

**Base case:** $10^{2(0)+1} + 1 = 10 + 1 = 11$, which is clearly divisible by $11$.

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**Induction hypothesis:** Assume that $(10^{2k+1} + 1)$ is divisible by $11$, we work to show that $(10^{2(k+1)+1} + 1)$ is divisible by $11$. That is, $(10^{2k+3} + 1) = (10^2*10^{2k+1} + 1) = $

Where am I going wrong?

- Solve $y^2= x^3 − 33$ in integers
- Finding all solutions of $x^{11}\equiv 1\bmod23,$
- Quadratics and divisibility
- $\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_k}$ not an integer
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- How come if $\ i\ $ not of the following form, then $12i + 5$ must be prime?
- Proof of an alternative form of Fermat-Euler's theorem.

Hint: Notice that $10^k=(-1)^k \pmod{11}$. So, for $k$ odd and $k’$ even, $10^{k} + 10^{k’}=0\pmod{11}$.

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